If $N$ and $G/N$ are virtually solvable, then $G$ is virtually solvable? ("Virtually solvable" means there's a solvable subgroup of finite index.)
I know this statement holds if "virtually solvable" is replaced by "solvable", but I want to knock it down to the finite-index subgroup case. I'm pretty sure this is a true statement, although you might have to add a "sub-exponential growth" hypothesis. Probably not though.
My line of attack is the same as in the proof of "$N$, $G/N$ solvable $\implies$ $G$ solvable". First, since $G/N$ is solvable, it has a subgroup $H/N$ of finite index and $H/N$ is solvable. Apply the correspondence theorem: $[G : H]  = [G/N : H/N] < \infty$, and a solvable series for $H/N$ corresponds to a chain of subgroups
$ H = H_0 \geq H_1 \geq \cdots \geq H_{m-1} \geq H_m = N $
where $H_k/H_{k+1}$ is abelian.
Next, $N$ has a subgroup $N_0$ with $[N : N_0] < \infty$ and $N_0$ is solvable. Then
$[G:N_0] = [G : H][H:N][N:N_0]$
and since $[G:H], [N:N_0] < \infty$ it suffices to show that $[H:N]<\infty$.
I'm stuck here. Is this the right direction to go with this? It doesn't seem like the above chain of subgroups $H_0,\ldots,H_m$ helps at all.

Yes. 
1/ Subcase: $N$ finite, $G/N$ solvable. Then the centralizer $C$ of $N$ has finite index (just because $N$ is finite). Since $C\cap N$ is abelian and the projection of $C$ in $G/N$ is solvable, it follows that $C$ is solvable, hence $G$ is virtually solvable. 
2/ Subcase: $N$ virtually solvable, $G/N$ solvable. Let $M$ be a finite index subgroup of $N$ if minimal index. Then $M$ is normal in $G$ (otherwise, if $M'$ is another conjugate of $M$, $MM'$ is solvable and of smaller index, as pointed out by D. Holt). By subcase 1, $G/M$ is solvable; since $M$ is solvable it follows that $G$ is solvable.
3/ General case: immediately reduces to the previous one.
