Let $K=SO(2,\mathbb{R}), \mathfrak{g}=gl(2,\mathbb{R})$. Let $V$ be a finite dimensional irreducible $(\mathfrak{g},K)$-module, W be a $(\mathfrak{g},K)$ module. I want to prove that $Ext_{(\mathfrak{g},K)}^{n}(\mathbb{C},Hom(V,W))=Ext_{(\mathfrak{g},K)}^n(V,W)$ for all $n$. Note $\mathbb{C}$ denotes the trivial $(\mathfrak{g},K)$-module.
Is such a statement true in greater generality? Something like $Ext_R^n(A,Hom(B,C))=Ext_R^n(A \otimes B,C)$? I know it's true for $n=0$ since $\_ \otimes B$ is left adjoint to $Hom(B,\_)$.
$\def\Ext{\mathop{\mathrm{Ext}}} \def\Hom{\mathop{\mathrm{Hom}}}$
The answer is yes, because you're taking the right derived functors of the "same" functor. Let $F$ be the functor $A \mapsto \Hom(A, \Hom(B, C))$ and $G$ is the functor $A \mapsto \Hom(A \otimes B, C).$ The adjunction between tensor and hom tells us that $F$ and $G$ are naturally isomorphic. Let $\eta$ be a natural isomorphism from $F$ to $G$.
Let $C$ be any object in your abelian category and let
$$\cdots \rightarrow P_{1} \rightarrow P_{0} \rightarrow C \rightarrow 0$$
be a projective resolution for $C$. Then, to get the ext groups on the left, you take the cohomology of the complex
$$0 \rightarrow F(P_{0}) \rightarrow F(P_{1}) \rightarrow \cdots$$
and to get the ext groups on the right, you take the cohomology of the complex
$$0 \rightarrow G(P_{0}) \rightarrow G(P_{1}) \rightarrow \cdots$$
But you can define an isomorphism $\eta_{P_{i}}$ from $F(P_{i}) \rightarrow G(P_{i})$ and naturality of $\eta$ tells us that this gives us a chain map. In particular, since $\eta_{P_{i}}$ is an isomorphism for each $i$, $\eta$ defines a chain homotopy and hence an isomorphism on homology groups.