$\newcommand\R{\mathbb R}\newcommand\N{\mathbb N}$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 $a$ 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 $G(a):=g^{(k)}(a)/k!$. Then, by a Taylor formula, 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*}
if $k\ge1$, with the similar formula for $G(x)$. 
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 $f^{(k)}(a)\ne0$.  

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**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 $g_k'(a)=0$. 

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**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.