Limit $\lim_{n\to\infty}\sum_{k=0}^n\frac{n^{2k}}{(k!)^2}\big/\sum_{k=0}^\infty\frac{n^{2k}}{(k!)^2}$
Here are two possible approaches:
Method 1. Let $X_n$ be a random variable with
$$ \mathbb{P}(X_n = k) = \frac{n^{2k}}{(k!)^2} \bigg/\biggl( \sum_{l=0}^{\infty} \frac{n^{2l}}{(l!)^2} \biggr), \qquad k = 0, 1, 2, \cdots. $$
Then the characteristic function of $X_n$ is given by
$$ \varphi_{X_n}(t) = \mathbb{E}[e^{it X_n}] = \frac{I_0(2n e^{it/2})}{I_0(2n)}, $$
where $I_0$ is the modified Bessel function of the first kind and order $0$. Now we normalize $X_n$ as follows:
$$ Z_n = \frac{X_n - n}{\sqrt{n}}. $$
Then by invoking the asymptotic formula for $I_0$:
$$ I_0(z) \sim \frac{e^{z}}{\sqrt{2\pi z}} \qquad \text{as} \quad z \to \infty \quad\text{along}\quad |\arg(z)| \leq \frac{\pi}{2}-\delta, $$
for each fixed $t \in \mathbb{R}$ it follows that
$$ \varphi_{Z_n}(t) = e^{-it\sqrt{n}} \, \frac{I_0(2n\exp(it/2\sqrt{n}))}{I_0(2n)} \sim \exp\bigl( 2ne^{it/2\sqrt{n}}-2n-it\sqrt{n} \bigr) \qquad \text{as} \quad n\to\infty. $$
This shows that
$$ \lim_{n\to\infty} \varphi_{Z_n}(t) = e^{-t^2/4}, $$
and so, $Z_n$ converges in distribution to $Z \sim \mathcal{N}(0, \frac{1}{2})$. Then the desired limit is
$$ \mathbb{P}(X_n \leq n) = \mathbb{P}(Z_n \leq 0) \xrightarrow[]{n\to\infty} \mathbb{P}(Z \leq 0) = \frac{1}{2}. $$
The second question seems also interesting and I suspect that it may be related to the local CLT, although I have no good idea in this direction.
Method 2. Here is a sketch of the proof using the Laplace's method:
By approximating the sum by integral and invoking the Stirling's formula, for any fixed large $N_0$ and for any $N \in \{N_0+1, N_0+2, \cdots\} \cup \{+\infty\}$, we expect:
$$ \sum_{n=N_0}^{N} \frac{n^{2k}}{(k!)^2} \approx \frac{1}{2\pi} \int_{N_0}^{N} \frac{n^{2x}}{x^{2x+1} e^{-2x}} \, \mathrm{d}x. $$
Now by writing
$$ \frac{n^{2x}}{x^{2x+1} e^{-2x}} = \exp\biggl( 2n - \log n - \frac{x-n}{n} - \int_{n}^{x} (x - t)\frac{2t-1}{t^2} \, \mathrm{d}t \biggr) $$
and substituting $x = n+\sqrt{n}z$ and $t = n+\sqrt{n}u$, we get
$$ \frac{1}{2\pi} \int_{N_0}^{N} \frac{n^{2x}}{x^{2x+1} e^{-2x}} \, \mathrm{d}x = \frac{e^{2n}}{\sqrt{2\pi n}} \int_{\frac{N_0-n}{\sqrt{n}}}^{\frac{N-n}{\sqrt{n}}} \exp\biggl( -\frac{z}{\sqrt{n}} - \int_{0}^{z} (z - u) \frac{2 + \frac{2u}{\sqrt{n}}-\frac{1}{n}}{\bigl( 1 + \frac{u}{\sqrt{n}}\bigr)^2} \, \mathrm{d}u \biggr) \, \mathrm{d}z. $$
Then, as $n\to\infty$, we expect this to become close to:
$$ \approx \frac{e^{2n}}{\sqrt{2\pi n}} \int_{\frac{N_0-n}{\sqrt{n}}}^{\frac{N-n}{\sqrt{n}}} \exp\biggl( - \int_{0}^{z} 2(z - u) \, \mathrm{d}u \biggr) \, \mathrm{d}z = \frac{e^{2n}}{\sqrt{2\pi n}} \int_{\frac{N_0-n}{\sqrt{n}}}^{\frac{N-n}{\sqrt{n}}} e^{-z^2} \, \mathrm{d}z. $$
Applying this to $N = n$ and $N = +\infty$ would then show that their ratio converges to
$$ \frac{\int_{-\infty}^{0} e^{-z^2} \, \mathrm{d}z}{\int_{-\infty}^{\infty} e^{-z^2} \, \mathrm{d}z} = \frac{1}{2}. $$
Addendum. For the second question, a numerical evidence suggests that
$$ \lim_{n\to\infty} \sqrt{n}\Biggl( \frac{\sum_{k=0}^{n} n^{2k}/(k!)^2}{\sum_{k=0}^{\infty} n^{2k}/(k!)^2} - \frac{1}{2} \Biggr) = \frac{5}{6\sqrt{\pi}}. $$
However, I have no simple idea for proving this.
Preliminaries
Lemma $\bf{1}$: For $-1\lt x\lt1$, $$ e^{-\frac{x}{1-x}}\le1-x\le e^{-x}\le\tfrac1{1+x}\le e^{-\frac{x}{1+x}}\tag{1} $$ Proof: For all $x\in\mathbb{R}$, Bernoulli's Inequality gives $$ \begin{align} 1+x &\le\lim_{n\to\infty}\left(1+\frac xn\right)^n\\ &=e^x\tag{1a} \end{align} $$ Taking the reciprocal of $\text{(1a)}$, for $x\gt-1$, gives $$ e^{-x}\le\frac1{1+x}\tag{1b} $$ Substituting $x\mapsto-x$ in $\text{(1a)}$ gives $$ 1-x\le e^{-x}\tag{1c} $$ Substituting $x\mapsto\frac{x}{1+x}$ in $\text{(1c)}$ gives $$ \frac1{1+x}\le e^{-\frac{x}{1+x}}\tag{1d} $$ Substituting $x\mapsto\frac{x}{1-x}$ in $\text{(1b)}$ gives, for $x\lt1$, $$ e^{-\frac{x}{1-x}}\le1-x\tag{1e} $$ $\large\square$
Lemma $\bf{2}$: For $|x-y|\le1$,
$$
\left|\,e^x-e^x\,\right|\le3|x-y|\,e^{\min(x,y)}\tag2
$$
Proof:
$$
\begin{align}
\left|\,e^x-e^y\,\right|
&\le|x-y|\,e^{\max(x,y)}\tag{2a}\\[3pt]
&=|x-y|e^{|x-y|}e^{\min(x,y)}\tag{2b}\\[3pt]
&\le3|x-y|\,e^{\min(x,y)}\tag{2c}
\end{align}
$$
Explanation:
$\text{(2a)}$: Mean Value Theorem
$\text{(2b)}$: $\max(x,y)=\min(x,y)+|x-y|$
$\text{(2c)}$: $e^{|x-y|}\lt3$ for $|x-y|\le1$
$\large\square$
Theorem $\bf{1}$: If $k\le n$ $$ e^{-\frac{k(k-1)}{2(n-k+1)}}\le\overbrace{\prod_{j=0}^{k-1}\left(1-\frac jn\right)}^{n^{\underline{k}}/n^k}\le e^{-\frac{k(k-1)}{2n}}\le\overbrace{\prod_{j=0}^{k-1}\left(1+\frac jn\right)^{-1}}^{n^k/n^{\overline{k}}}\le e^{-\frac{k(k-1)}{2(n+k-1)}}\tag3 $$ Proof: Set $x=\frac jn$ in Lemma $1$: $$ e^{-\frac{j}{n-j}}\le1-\frac{j}{n}\le e^{-\frac{j}{n}}\le\frac1{1+\frac{j}{n}}\le e^{-\frac{j}{n+j}}\tag{3a} $$ For $0\le j\le k-1$, $\text{(3a)}$ implies $$ e^{-\frac{j}{n-k+1}}\le1-\frac{j}{n}\le e^{-\frac{j}{n}}\le\frac1{1+\frac{j}{n}}\le e^{-\frac{j}{n+k-1}}\tag{3b} $$ Take the product of $\text{(3b)}$ from $j=0$ to $j=k-1$.
$\large\square$
Inequality $\bf{1}$: If $k\le n^{5/9}$, then
$$
\begin{align}
\frac{k(k-1)^2}{n^2-(k-1)^2}
&\le\frac{n^{5/9}\left(n^{5/9}-1\right)^2}{n^2-\left(n^{5/9}-1\right)^2}\tag{4a}\\
&\le\frac{n^{10/9}\left(n^{5/9}-1\right)}{n^2-n^{13/9}}\tag{4b}\\[3pt]
&=\frac1{n^{1/3}}\tag{4c}
\end{align}
$$
Explanation:
$\text{(4a)}$: $k\le n^{5/9}$
$\text{(4b)}$: $n^{5/9}-1\le n^{5/9}$ and $\left(n^{5/9}-1\right)^2\le n^{13/9}$
$\text{(4c)}$: cancel common factors
Inequality $\bf{2}$: If $k\gt n^{5/9}$, then
$$
\begin{align}
\frac{k(k-1)}{n+k-1}
&\ge\frac{k(k-1)}{k^{9/5}+k-1}\tag{5a}\\
&\ge k^{1/5}-2k^{-3/5}\tag{5b}\\
&\ge k^{1/5}-\frac2{n^{1/3}}\tag{5c}
\end{align}
$$
Explanation:
$\text{(5a)}$: $n\lt k^{9/5}$
$\text{(5b)}$: cross multiply and compare
$\text{(5c)}$: $k\gt n^{5/9}$
Approximating the squares of $\boldsymbol{n^k/n^{\overline{k}}}$ and $\boldsymbol{n^{\underline{k}}/n^k}$
Choose $\epsilon\gt0$ and let $n\ge\max\!\left(\epsilon^{-3},8\right)$.
If $k\le n^{5/9}$, then
$$
\begin{align}
\left|\,\left(\frac{n^k}{n^{\overline{k}}}\right)^2-e^{-\frac{k(k-1)}{n}}\,\right|
+\left|\,e^{-\frac{k(k-1)}{n}}-\left(\frac{n^{\underline{k}}}{n^k}\right)^2\,\right|
&\le\left|\,e^{-\frac{k(k-1)}{n+k-1}}-e^{-\frac{k(k-1)}{n-k+1}}\,\right|\tag{6a}\\
&\le3\frac{2k(k-1)^2}{n^2-(k-1)^2}\,e^{-\frac{k(k-1)}{n}}\tag{6b}\\[6pt]
&\le6\epsilon\,e^{-\frac{k(k-1)}{n}}\tag{6c}
\end{align}
$$
Explanation:
$\text{(6a)}$: Theorem $1$
$\text{(6b)}$: Lemma $2$
$\text{(6c)}$: Inequality $1$ implies $\frac{2k(k-1)^2}{n^2-(k-1)^2}\le\min(1,2\epsilon)$
If $k\gt n^{5/9}$, then Inequality $2$ says then $$ \frac{k(k-1)}{n-k+1}\ge\frac{k(k-1)}{n}\ge\frac{k(k-1)}{n+k-1}\ge k^{1/5}-1\tag7 $$ Thus, the squares of the remainders outside of the range where $(6)$ holds can be bounded by $$ \sum_{k\gt n^{5/9}}e^{-k^{1/5}+1}=O\!\left(n^{4/9}e^{-n^{1/9}}\right)\tag8 $$ Furthermore, using Riemann Sums, we have $$ \begin{align} \frac1{\sqrt{n}}\sum_{k=0}^n e^{-\frac{k(k-1)}{n}} &=\int_0^\infty e^{-x^2}\,\mathrm{d}x+O\!\left(\frac1{\sqrt{n}}\right)\\ &=\frac{\sqrt\pi}2+O\!\left(\frac1{\sqrt{n}}\right)\tag9 \end{align} $$ since the variation of $e^{-x^2}$ is $1$ and the step-size is $\frac1{\sqrt{n}}$.
Answer to Part $\bf{1}$
Computing the sum for $\boldsymbol{m\le n}$:
$$
\begin{align}
\sum_{m=0}^n\left(\frac{n^m}{m!}\right)^2
&=\sum_{k=0}^n\left(\frac{n^{n-k}}{(n-k)!}\right)^2\tag{10a}\\
&=\left(\frac{n^n}{n!}\right)^2\sum_{k=0}^n\left(\frac{n^{\underline{k}}}{n^k}\right)^2\tag{10b}\\
&=\left(\frac{n^n}{n!}\right)^2\left[\frac{\sqrt{\pi n}}2+O\!\left(n^{1/6}\right)\right]\tag{10c}
\end{align}
$$
Explanation:
$\text{(10a)}$: $m=n-k$
$\text{(10b)}$: pull out a common factor
$\text{(10c)}$: $(6)$ and $(8)$ say that $\sum\limits_{k=0}^n\left(\frac{n^{\underline{k}}}{n^k}\right)^2=\sum\limits_{k=0}^ne^{-\frac{k(k-1)}{n}}\left(1+O\!\left(n^{-1/3}\right)\right)+O\!\left(n^{4/9}e^{-n^{1/9}}\right)$
$\phantom{\text{(10c):}}$ which, by $(9)$, is $\frac{\sqrt{\pi n}}2+O\!\left(n^{1/6}\right)$
Computing the sum for $\boldsymbol{m\gt n}$:
$$
\begin{align}
\sum_{m=n+1}^\infty\left(\frac{n^m}{m!}\right)^2
&=\sum_{k=2}^\infty\left(\frac{n^{n+k-1}}{(n+k-1)!}\right)^2\tag{11a}\\
&=\left(\frac{n^n}{n!}\right)^2\sum_{k=2}^\infty\left(\frac{n^k}{n^{\overline{k}}}\right)^2\tag{11b}\\
&=\left(\frac{n^n}{n!}\right)^2\left[\sum_{k=0}^\infty\left(\frac{n^k}{n^{\overline{k}}}\right)^2-2\right]\tag{11c}\\
&=\left(\frac{n^n}{n!}\right)^2\left[\frac{\sqrt{\pi n}}2+O\!\left(n^{1/6}\right)\right]\tag{11d}
\end{align}
$$
Explanation:
$\text{(11a)}$: $m=n+k-1$
$\text{(11b)}$: pull out a common factor
$\text{(11c)}$: $n^k/n^{\overline{k}}=1$ for $k=0$ and $k=1$
$\text{(11d)}$: $(6)$ and $(8)$ say that $\sum\limits_{k=0}^\infty\left(\frac{n^k}{n^{\overline{k}}}\right)^2=\sum\limits_{k=0}^ne^{-\frac{k(k-1)}{n}}\left(1+O\!\left(n^{-1/3}\right)\right)+O\!\left(n^{4/9}e^{-n^{1/9}}\right)$
$\phantom{\text{(11d):}}$ which, by $(9)$, is $\frac{\sqrt{\pi n}}2+O\!\left(n^{1/6}\right)$
Thus, $(10)$ and $(11)$ imply $$ \bbox[5px,border:2px solid #C0A000]{\quad\frac{\displaystyle\sum\limits_{m=0}^n\left(\frac{n^m}{m!}\right)^2}{\displaystyle\sum\limits_{m=0}^\infty\left(\frac{n^m}{m!}\right)^2}=\frac12+O\!\left(n^{-1/3}\right)\quad}\tag{12} $$ An error term of $O\!\left(n^{-1/3}\right)$ is insufficient to get the answer to Part $2$.
More Preliminaries
Squaring the two leftmost inequalities from $(3)$:
$$
e^{-\frac{k^2-k}{n-k+1}}\le\prod_{j=0}^{k-1}\left(1-\frac{j}{n}\right)^2\le e^{-\frac{k^2-k}{n}}\tag{13}
$$
Similar to Theorem $1$, but setting $x=\frac{j^2}{n^2}$,
$$
\begin{align}
e^{\frac{2k^3-3k^2+k}{3n^2}}-1&\le\prod_{j=0}^{k-1}\left(1-\frac{j^2}{n^2}\right)^{-2}-1\le e^{\frac{2k^3-3k^2+k}{3n^2-3(k-1)^2}}-1\tag{14a}\\
\frac{2k^3-3k^2+k}{3n^2}&\le\prod_{j=0}^{k-1}\left(1-\frac{j^2}{n^2}\right)^{-2}-1\le\frac{2k^3-3k^2+k}{3n^2-2k^3+5k-1}\tag{14b}
\end{align}
$$
Explanation:
$\text{(14a})$: $\sum\limits_{j=0}^{k-1}j^2=\frac{2k^3-3k^2+1}6$
$\text{(14b})$: $x\le e^x-1$ and $e^x-1\le\frac{x}{1-x}$
For $k\le n^{5/9}$, $(13)$ is $e^{-\frac{k^2}n}\left(1+O\!\left(n^{-1/3}\right)\right)$ and $(14)$ is $\frac{2k^3}{3n^2}\left(1+O\!\left(n^{-1/3}\right)\right)$.
For $k\gt n^{5/9}$, the bounds of $(8)$ still hold.
Using Riemann Sums we have $$ \begin{align} \sum_{k=0}^\infty\frac{2k^3}{3n^2}e^{-\frac{k^2}{n}} &=\frac23\int_0^\infty x^3e^{-x^2}\,\mathrm{d}x\tag{15a}+O\!\left(\frac1{\sqrt{n}}\right)\\ &=\frac13+O\!\left(\frac1{\sqrt{n}}\right)\tag{15b} \end{align} $$ because the variation of $x^3e^{-x^2}$ is $\sqrt{\frac{27}2}e^{-3/2}$ and the step size is $\frac1{\sqrt{n}}$.
Approximating the difference of the squares of $\boldsymbol{n^k/n^{\overline{k}}}$ and $\boldsymbol{n^{\underline{k}}/n^k}$
$$
\begin{align}
\left(\frac{n^k}{n^{\overline{k}}}\right)^2-\left(\frac{n^{\underline{k}}}{n^k}\right)^2
&=\prod_{j=0}^{k-1}\left(1+\frac jn\right)^{-2}-\prod_{j=0}^{k-1}\left(1-\frac jn\right)^2\tag{16a}\\
&=\prod_{j=0}^{k-1}\left(1-\frac jn\right)^2\left(1-\frac{j^2}{n^2}\right)^{-2}-\prod_{j=0}^{k-1}\left(1-\frac jn\right)^2\tag{16b}\\
&=\left[\prod_{j=0}^{k-1}\left(1-\frac{j^2}{n^2}\right)^{-2}-1\right]\prod_{j=0}^{k-1}\left(1-\frac jn\right)^2\tag{16c}\\[3pt]
&=\frac{2k^3}{3n^2}\,e^{-\frac{k^2}{n}}\left(1+O\!\left(n^{-1/3}\right)\right)\tag{16d}
\end{align}
$$
Explanation:
$\text{(16a)}$: write the fractions as products
$\text{(16b)}$: $(1+x)^{-1}=(1-x)\left(1-x^2\right)^{-1}$
$\text{(16c)}$: redistribute a common factor
$\text{(16d)}$: $(13)$ and $(14)$
Answer to Part $\bf{2}$
$$
\begin{align}
\frac{\displaystyle\sum_{m=0}^n\left(\frac{n^m}{m!}\right)^2}{\displaystyle\sum_{m=0}^\infty\left(\frac{n^m}{m!}\right)^2}-\frac12
&=\frac12\,\frac{\displaystyle\sum_{m=0}^n\left(\frac{n^m}{m!}\right)^2-\sum_{m=n+1}^\infty\left(\frac{n^m}{m!}\right)^2}{\displaystyle\sum_{m=0}^n\left(\frac{n^m}{m!}\right)^2+\sum_{m=n+1}^\infty\left(\frac{n^m}{m!}\right)^2}\tag{17a}\\
&=\frac12\,\frac{\displaystyle\sum_{k=0}^n\left(\frac{n^{\underline{k}}}{n^k}\right)^2-\sum_{k=0}^\infty\left(\frac{n^k}{n^{\overline{k}}}\right)^2+2}{\displaystyle\sum_{k=0}^n\left(\frac{n^{\underline{k}}}{n^k}\right)^2+\sum_{k=0}^\infty\left(\frac{n^k}{n^{\overline{k}}}\right)^2-2}\tag{17b}\\
&=\frac12\,\frac{\displaystyle2-\sum\limits_{k=0}^\infty\frac{2k^3}{3n^2}\,e^{-\frac{k^2}{n}}\left(1+O\!\left(n^{-1/3}\right)\right)}{\sqrt{\pi n}+O\!\left(n^{1/6}\right)}\tag{17c}\\
&=\frac12\,\frac{\displaystyle\frac53+O\!\left(n^{-1/3}\right)}{\sqrt{\pi n}+O\!\left(n^{1/6}\right)}\tag{17d}\\[9pt]
&=\frac5{6\sqrt{\pi n}}+O\!\left(n^{-5/6}\right)\tag{17e}
\end{align}
$$
Explanation:
$\text{(17a)}$: split the sum in the denominator into two parts
$\text{(17b)}$: apply equations $\text{(10b)}$ and $\text{(11c)}$ and cancel the factors of $\left(\frac{n^n}{n!}\right)^2$
$\text{(17c)}$: apply $(16)$ and $\text{(10c)}$ and $\text{(11d)}$
$\text{(17d)}$: apply $(15)$
$\text{(17e)}$: simplify
Therefore, $$ \bbox[5px,border:2px solid #C0A000]{\quad\sqrt{n}\left[\frac{\displaystyle\sum_{m=0}^n\left(\frac{n^m}{m!}\right)^2}{\displaystyle\sum_{m=0}^\infty\left(\frac{n^m}{m!}\right)^2}-\frac12\right] =\frac5{6\sqrt\pi}+O\!\left(n^{-1/3}\right)\quad}\tag{18} $$