Let $A$ be a commutative ring, $I$ an ideal of $A$ and $M$ a module over $A$. Is it true that $Ann(M/IM) = Ann(M) + I$? One inclusion is certainly true, but I don't know about the other one. If the above statement does not hold in general, does it perhaps for finitely generated modules?
Replace $A$ with $A/Ann(M)$. Then $M$ is an $A$-module with trivial annihilator ideal. Let $a\in Ann(M/IM)$ (equivalent to (EDIT) $(aM+IM)/IM=0$). You would like to conclude that $a\in I$, or equivalently, that $(aA+I)/I=0$.
This is true by definition if $M$ is faithfully flat over $A$ (e.g. if $M$ is free of positive rank over $A$) EDIT because then $((aA+I)/I)\otimes_A M $ is isomorphic to $(aM+IM)/IM=0$ by flatness, hence $(aA+I)/I=0$ by faithful flatness. Otherwise it is false even when $M$ is finitely generated over $A$.
Example: let $A=k[x,y]\subset k[t]$ where $k$ is a field, $x=t^2, y=t^3$. Let $M=k[t]$. It is finitely generated over $A$ (a system of generators is $1, t$). Let $I=xA$. Then $IM=t^2k[t]$ and $yM=t^3k[t]\subseteq IM$. So $y\in Ann(M/IM)$. But $y\notin I$.