Let $G$ be a group and $H$ a subgroup of the center of the group $G$ (so $H$ is normal in $G$). Suppose that $G/H$ is a residually finite group, then is it true that $G$ is a residually finite group?
Moreover, what can we say about the residual finiteness of the inner automorphism group of $G$ when $H$ is a proper subgroup of the center?
No it's not true: while f.g. metabelian groups are all residually finite, there are center-by-metabelian f.g. groups that are not residually finite. Here's an example due to Ph. Hall (multi-recyclable to many group-theoretic questions).
Write $$M(t,x,y,z)=\begin{pmatrix} 1 & x & z\\ 0 & t & y\\ 0 & 0 & 1\end{pmatrix},\quad d(t)=\begin{pmatrix} t & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{pmatrix},\quad \zeta(z)=\begin{pmatrix} 1 & 0 & z\\ 0 & 1 & 0\\ 0 & 0 & 1\end{pmatrix}.$$
Fix a prime $p$. Let $G_0$ be the group of matrices $M(p^n,x,y,z)$ when $n$ ranges over $\mathbf{Z}$ and $(x,y,z)$ over $\mathbf{Z}[1/p]^3$. It's finitely generated, and its center $Z$ consists of those $\zeta(z)$ for $z\in\mathbf{Z}[1/p]$. Let $Z'$ be the central subgroup of those $\zeta(z)$ for $z\in\mathbf{Z}$ and $G=G_0/Z'$. Its center is reduced to $Z/Z'$.
Conjugation by the matrix $d(p)$ induces a non-injective surjective endomorphism of $G$, which is therefore not Hopfian, hence not residually finite (which is also seen by observing that the Prüfer group $Z/Z'$ is not residually finite).
But the central quotient $G/(Z/Z')=G_0/Z$ is metabelian (it's a semidirect product $\mathbf{Z}\ltimes\mathbf{Z}[1/p]^2$, action by multiplication by $(p,p^{-1})$), hence is residually finite.
Other examples:
some come from lattices in Lie groups. E.g., the inverse image of $\mathrm{SL}_3(\mathbf{Z}[1/2])$ (residually finite) in the universal (2-fold) covering of $\mathbf{SL}_3(\mathbf{R})$ is, I believe, not residually finite. I'm not sure of a reference, but Deligne and Raghunathan produced others using other arithmetic lattices, and Erschler did a similar game using Grigorchuk's first group.