Convolution of an $L^p$ function with a Compactly supported continuous function. What does it give?
WARNING. This answer is correct only in dimension $n=1$.
The answer to Q2 (the $p=\infty$ case) is negative; a counterexample is $f=1$ and $g$ such that $\hat{g}(0)=0$. Indeed, $$ 1\ast g(x)=\int_{\mathbb R^n} g(y)\, dy=\hat{g}(0)=0.$$
The answer to Q1 is affirmative.
EDIT. I just found a mistake. The following is correct only for $n=1$. I still think that the result is true in generality, but it needs some more work.
If $f\ast g=0$, then $\hat{f}\hat{g}=0$; note that this last product makes sense, as $\hat{g}$ is a smooth function, and actually it is analytic, while $\hat{f}$ is a tempered distribution.
Now, $\hat{g}$ only has isolated zeros, by analyticity; denote them by $z_1, z_2, \ldots$.
EDIT. The above needs not be true for $n>1$. For example, the compactly supported function $g(x)=\prod_{j=1}^n \mathbf 1_{[-1, 1]}(x_j)$ has the Fourier transform $$ \hat{g}(\xi)=\prod_{j=1}^n \frac{\sin \xi_j}{\xi_j}, $$ which is analytic and which has non-isolated zeros. END EDIT.
Thus, $\hat{f}\hat{g}=0$ implies that $\hat{f}$ is a distribution supported in $\{z_1, z_2, \ldots\}$, and the only such distributions have the form $$ \hat{f}=\sum_{j=1}^\infty \sum_{m\in N(j)} a_{mj} \partial^m \delta_{z_j}, $$ where $N(j)$ is a finite set of multi-indices, and $a_{mj}\in \mathbb C$. $^{[1]}$
Therefore, inverting the Fourier transform, we see that $$ f=\sum_{j=1}^\infty \sum_{m\in N(j)} a_{mj} i^{|m|} e^{i x\cdot z_j} x^m, $$ which is a contradiction with the assumption $f\in L^p(\mathbb R^n)$, with $p<\infty$.
$^{[1]}$ A multi-index $m$ is a $n$-uple $(m_1, m_2, \ldots, m_n)\in\mathbb N^n$. We denote $\partial^m:=\partial_{x_1}^{m_1}\ldots \partial_{x_n}^{m_n}$ and $x^m :=x_1^{m_1}\ldots x_n^{m_n}$ and $\lvert m\rvert:= m_1+\ldots + m_n$. This is a standard terminology of distribution theory.
The result is not true for $n\geq 2$, at least not in the full range: Let $p_0=2n/(n-1)$, then for every $p>p_0$ there exists $f \in L^p(\mathbb{R}^n)$ and $g\in C_c(\mathbb{R}^n)$ such that $g\neq 0 \neq f$ and $f*g=0$.
We'll take $f= \check{\sigma}$, where $\sigma= \mathcal{H}^{n-1}|_{\mathbb{S}^{n-1}}$ is the surface measure on the sphere. It's known (see for example Grafakos's book "Modern Fourier Analysis", it's an appendix if I recall correctly) $$ |f(x)|\lesssim (1+|x|)^{-(n-1)/2}. $$ Therefore $f\in L^p(\mathbb{R}^n)$ for every $p>p_0$.
Now consider the function $H:\mathbb{C}^n\to \mathbb{C}$ given by $$ H(z)=e^{i(z_1+\ldots + z_n)}(z_1^2+\ldots +z_n^2-1), $$ where $z=(z_1,\ldots,z_n)$. Then $H$ is entire and $$ |H(z)|\leq (1+|z|)^2 e^{|\text{Im}(z)|}, $$ so that by a Paley-Wiener Theorem (see Strichartz's book on distributions, or the Wikipedia page) $H=\hat{h}$ for a compactly supported distribution $h$. Now take $g=h*\eta$ for some $\eta\in C_c^\infty(\mathbb{R}^n)$ to obtain that $g\in C_c(\mathbb{R}^n)$ and $\hat{g}= H \hat{\eta}$, which vanishes on the unit sphere.
Note that for $n=1$ we have no decay since the measure $\sigma$ is just a sum of Dirac masses. This agrees with @GiuseppeNegro's result that $p=\infty$ is the only failure there. On the other hand for $p>2$ and $n\geq 2$ Fourier transforms of $L^p$ functions may be supported on lower dimensional sets. I'm not sure that $p_0$ is sharp though, so one question remains:
Is the original question true for $p\in (2,2n/(n-1)]$?