This is exercise 1.2.27 of Bruns-Herzog:
Let $R$ be a Noetherian ring, $M$ a finite $R$-module and $N$ an arbitrary $R$-module. Deduce that $\operatorname{Ass}(\operatorname{Hom}_R(M,N)) = \operatorname{Ass}N \cap \operatorname{Supp}M$. (answered here )
I want to know if there is a similar expression for $\operatorname{Ass}\operatorname{Ext}^i(M,N)$ (with extra assumptions if needed)? Specially I want to know $\operatorname{Ass}\operatorname{Ext}^i(R/I,N)$, when $R$ is local.
I know that $\operatorname{Ass}\operatorname{Ext}^i(R/I,N) \subseteq \operatorname{supp}\operatorname{Ext}^i(R/I,N) \subseteq \operatorname{supp}N \cap \operatorname{supp}R/I$, but is there a kind of equality?