$\text{Ext}_R^n(\bigoplus_{i\in I} M_i, N) = \prod_{i\in I}\text{Ext}_R^n(M_i,N)$ Let $(M_i)_{i\in I}$ be a collection of $R$-moduls. Show that for all $N\in \text{Ob}(_R\text{Mod})$ is 
$$
\text{Ext}_R^n(\bigoplus_{i\in I} M_i, N) = \prod_{i\in I}\text{Ext}_R^n(M_i,N).
$$
My idea is to make use of the universal property of the direct sum (note that $N\rightarrow I^{\bullet}$ is an injective resolution of $N$):
$$\begin{eqnarray*}
\text{Ext}_R^n(\bigoplus_{i\in I}M_i,N) & = &\mathcal{H}^n(\text{Hom}_R(\bigoplus_{i\in I}, I^{\bullet}))\\
& \cong & \mathcal{H}^n(\prod_{i\in I}\text{Hom}_R(M_i, I^{\bullet}))\\
& \cong & \prod_{i\in I}\mathcal{H}^n(\text{Hom}_R(M_i,I^{\bullet}))\\
& = & \prod_{i\in I}\text{Ext}_R^n(M_i,N)
\end{eqnarray*}$$
Is this correct? I'm not sure if & why $\mathcal{H}^n(\prod_{i\in I}\text{Hom}_R(M_i, I^{\bullet})) \cong  \prod_{i\in I}\mathcal{H}^n(\text{Hom}_R(M_i,I^{\bullet}))$ holds....

The proof is correct. If $(C_i)_{i\in I}$ is any family of cochain complexes, then by writing out the definitions you immediately see $H^n(\prod C_i)=\prod H^n(C_i)$.
For if $D$ denotes the differential on $\prod C_i$, $d_i$ the differential on $C_i$, then
$H^n(\prod C_i)=Ker D/Im D=\prod Ker d_i/\prod Im d_I=\prod Ker d_i/ Im d_i=\prod H^n(C_i).$
