This question is a follow up to Sasha's comment in Duals and Tensor products.
In the comment there, it is claimed that the given a ring $A$ and modules $M$, $N$, there is an isomorphism $$ RHom(M,A)\otimes^L RHom(N,A) = RHom(M\otimes^L N, A). $$
Question 1. Shouldn't we make assumption on $A$ to make sure the RHom's are bounded above in order for the (derived) tensor product to exist?
Then it is claimed that if $M$ is finite projective, then $RHom(M,A) = Hom(M,A)$ (obvious) and that the zero-th homology group of $Hom(M,A)\otimes^L RHom(N,A)$ is $Hom(M,A)\otimes Hom(N,A)$...
Question 2: Why is this last statement true? It would be true if $Hom(M,A)$ is flat, but why is it true in general?