Suppose that $(A,m)$ is a Noetherian local ring, $M$ is an $A$finite module. Assume that $x_1, ..., x_n$ are elements in $m$. Is the following equality true:
$$ \mbox{ann}(M/(x_1, ..., x_n)M) = (x_1, ..., x_n) + \mbox{ann}(M). $$
Suppose that $(A,m)$ is a Noetherian local ring, $M$ is an $A$finite module. Assume that $x_1, ..., x_n$ are elements in $m$. Is the following equality true: $$ \mbox{ann}(M/(x_1, ..., x_n)M) = (x_1, ..., x_n) + \mbox{ann}(M). $$ 


By modding out ${ann} (M)$ one can assume that $ann(M)=0$. Then the following is true: $$I \subseteq ann(M/IM) \subseteq \bar I $$ Here $\bar I$ denotes the integral closure of $I$. You can prove it using the determinantal trick (the one used in the proof of Nakayama's Lemma). In particular equality happens if $I$ is integrally closed. Now, a simple counterexample for the equality you wrote is $R=k[[x,y]]/(x^2y^3)$ and $M$ be the ideal $(x,y)$. Then you can check that $y^2M \subseteq xM$, thus $y^2 \in ann(M/xM)$. Note that $y^4yx^2=0$, so $y^2 \in \overline{(x)}$. 

