If $R$ is a non-Noetherian ring, $M$ is a finitely presented $R$-module, and $N$ is a finitely generated $R$-module, does it hold that Hom$_R(M,N)$ is a finitely generated $R$-module?   We tried applying the Hom$(-,N)$ functor to the sequence $R^n \rightarrow R^m \rightarrow M \rightarrow 0$ and using the fact that $N$ f.g. implies Hom$(R^m,N)$ are f.g., but we have no clue how to proceed.
Assume that $M$ and $N$ are two finitely generated $R$-modules. Then $\operatorname{Hom}_{R}(M,N)$ is a finitely generated $\mathbb Z$-module and/or $R$-module (in this case, assume that $R$ is commutative)?   Note that $R$ is not Noetherian ring. Is there any counterexample (for any cases)?
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