Let $M$ be an $R$-module and $I$ an $R$-ideal.
a) Show that if $M_m=0$ for every maximal ideal $m$ containing $I$, then $M=IM$.
b) Show that the converse holds in case $M$ is finite.
Proof of (a): Consider the $R$-module $\overline{M}=M/IM$ and let $m\subseteq R$ be a maximal ideal of $R$. If $I\not\subseteq m$, then $I_m=R_m$ and $\overline{M}_m=0$. If $I\subseteq m$ then by assumption $M_m=0$ and $\overline{M}_m=0$. Hence $\overline{M}=0$. i.e. $M=IM$.
Proof of (b): How would I do this part?
Suppose that $M$ is finitely generated, Let $L$ be a maximal ideal containing $I$, $LM=M$, by Nakayama lemma, there exists $l\in R$ which is not an element of $L$ such that $lM=0$. Since $M_L$ is finite dimensional vector space such that $lM_L=0$ where $l$ is a non zero element of the field $R/L$, you deduce that $M_L=0$.