Please give me a hint for this problem.
Let $R$ be a ring, and $I$ a right ideal of $R$. Show if $M$ is a left $R$-module, then
$$ f \ \colon (R/I) \otimes_{R} M \to M/IM$$
defined by $ f((r+I) \otimes m) = rm + IM$ is an isomorphism of commutative groups.
And show that if $L$ is a left ideal, then $$ (R/I) \otimes _{R} (R/L) \cong R/(I+L)$$ as commutative groups.
If we tensor with $M$ the short exact sequence $0\to I\to R\to R/I\to 0$ and consider the exact sequence $0\to IM\to M\to M/IM\to 0$, we obtain a commutative diagram of abelian groups with exact rows $$\require{AMScd} \begin{CD} {} @. I\otimes_RM @>>> R\otimes_RM @>>> (R/I)\otimes_R M @>>> 0 \\ @. @VaVV @VbVV @VcVV @. \\ 0 @>>> IM @>>> M @>>> M/IM @>>> 0 \end{CD} $$ We have $a(r\otimes x)=rx$, $b(r\otimes x)=rx$ and $c((r+I)\otimes x)=rx+IM$. Note that $a$ is surjective and $b$ is an isomorphism. A very simple diagram chasing shows that $c$ is an isomorphism as well.
For the second case, we need to compute $$ I(R/L)=(IR+L)/L=(I+L)/L $$ and so $$ (R/L)\big/I(R/L)=(R/L)\big/((I+L)/L)\cong R/(I+L) $$ by the homomorphism theorem.