Perhaps the best way to think about this is as follows: pick your favorite injective resolution for N and favorite projective resolution of M. Then $\mathrm{Ext}(M,N)$ is given by taking Hom between these complexes (NOT chain maps, just all maps of representations between the underlying modules), and putting a differential on those in the usual way.

Now, use the usual identification of $\mathrm{Hom}(A,B)\cong \mathrm{Hom}(A\otimes B^*,1)$ on this complex. So you see, it's the same as if we had tensored the projective resolution of $M$ with the dual of the injective resolution of $N$, which is a projective resolution of $N^*$, and then taken Hom to 1. Of course, the tensor product of two projective resolutions is a projective resolution of the tensor product, so we see this complex also computes $\mathrm{Hom}(N^*\otimes M,1)$.`

It also follows by abstract nonsense in one line: isomorphic functors have isomorphic derived functors.