$\newcommand\R{\mathbb R}\newcommand\N{\mathbb N}$Here are answers to your three questions (the latter two of them partial).
Answer 1: Yes, for any real $a$ and any $k\in\{0,1,\dots\}$,
\begin{equation*}
\text{if $h$ does not have all the derivatives at $a$, then $f^{(k)}(a)=0$.}\tag{1}
\end{equation*}
Indeed, say that $h$ is bad at $a$ (or, equivalently, that $a$ is a bad point of $h$0 if $h$ does not have all the derivatives at $a$.
Take any $a\in\R$. Suppose that $h$ is bad at $a$. Take then the smallest $k\in\{0,1,\dots\}$ such that $f^{(k)}(a)\ne0$, if such a $k$ exists. Let
\begin{equation*}
g:=fh
\end{equation*}
and
\begin{equation*}
F(x):=\frac{f(x)}{(x-a)^k}, \quad G(x):=\frac{g(x)}{(x-a)^k}
\end{equation*}
for real $x\ne a$, with $F(a):=f^{(k)}(a)/k!$ and $G(a):=g^{(k)}(a)/k!$. By a Taylor formula, if $k\ge1$, then for all real $x$
\begin{equation*}
F(x)=\frac1{(k-1)!}\int_0^1(1-s)^{k-1}f^{(k)}(a+(x-a)s)\,ds
\end{equation*}
and hence for all nonnegative integers $l$
\begin{equation*}
F^{(l)}(x)=\frac1{(k-1)!}\int_0^1(1-s)^{k-1}s^lf^{(k+l)}(a+(x-a)s)\,ds,
\end{equation*}
with the similar formulas for $G(x)$ and $G^{(l)}(x)$.
So, if $k\ge1$, then $F$ and $G$ are smooth, $F(a)\ne0$, and hence $F\ne0$ and $h=G/F$ on a neighborhood $V$ of $a$ (the equality $h(a)=G(a)/F(a)$ follows by continuity); the same conclusions obviously hold for $k=0$.
So, $h$ is smooth on $V$, which contradicts the assumption that $h$ is bad at $a$. So, as claimed, there is no $k\in\{0,1,\dots\}$ such that $f^{(k)}(a)\ne0$.
We have actually proved more than (1):
\begin{equation*}
\text{if $a$ is in the closure of the set of all bad points of $h$, } \\
\text{then $f^{(k)}(a)=0$ for all $k\in\{0,1,\dots\}$.}
\end{equation*}
Answer 2 (partial):
Now it follows that $g_k:=f^{(k)}h$ must be differentiable. Indeed, take any real $a$. If $h$ has all the derivatives at $a$, then so does $g_k$. If $h$ does not have all the derivatives at $a$, then, by (1), $f'(x)=o(|x-a|)$ as $x\to a$, so that $g_k(x)=o(|x-a|)$ as $x\to a$, so that $g_k'(a)=0$.
Answer 3 (partial):
Finally, let $g_{k,l}:=f^{(k)}h^{(l)}$ wherever $h^{(l)}$ exists, with $g_{k,l}:=0$ elsewhere. Take any real $a$ such that $h$ does not have all the derivatives at $a$. Note that $h'$ is bounded on the set where $h'$ exists, since $h$ is Lipschitz.
By (1), $f^{(k)}(x)=o(|x-a|)$ as $x\to a$, so that $g_{k,1}(x)=o(|x-a|)$ as $x\to a$, and hence $g_{k,1}'(a)=0$. So, $g_{k,1}$ is differentiable.
