Proving $f : (-1, 1) \rightarrow \mathbb{R}$ has a property

For the second question: Suppose the result fails. Then $f^{(k)}(0)\ne 0$ for some $k\in \{0,1,\dots,n-1\}.$ Let $k_0$ be the smallest such $k.$ We then have $f^{(k_0)}(0)\ne 0$ and $f^{(k)}(0)= 0$ for $k<k_0.$

By the theorem you cited, for each $x\in (-1,1)\setminus\{0\}$ there exists $z_x$ beween $0$ and $x$ such that

$$f(x) = \frac{f^{(k_0)}(z_x)}{k_0!}x^{k_0}.$$

It follows that for such $x.$

$$M|x|^n \ge \frac{|f^{(k_0)}(z_x)|}{k_0!}|x|^{k_0}.$$

Therefore

$$Mk_0!|x|^{n-k_0} \ge |f^{(k_0)}(z_x)|.$$

Now $n>k_0.$ Thus as $x\to 0,$ the left side $\to 0.$ Because $x\to 0$ forces $z_x\to 0,$ the right side $\to |f^{(k_0)}(0)|.$ The reason I can say the last is that $f^{(k_0)}$ is differentiable on $(-1,1),$ hence is continuous there.

We have thus shown $f^{(k_0)}(0)=0,$ contradiction. Thus there is no such $k_0,$ and this gives the desired conclusion.