The following addresses your second question in the case that $R$ is commutative, about $\text{Hom}_R(M,N)$.  First of all, it is false if there is not some additional hypothesis.  Let $R$ be the zero-dimensional, local, Noetherian, Gorenstein ring $k[x]/\langle x^2 \rangle$.  Let $M$ and $N$ both be $R/xR = k$.  Then $\text{Id}_M\in \text{Hom}_R(M,M)$ is not in the image of the morphism $\text{Hom}_R(M,R)\otimes_R M \to \text{Hom}_R(M,M)$.  The only reasonable additional hypothesis I can see is that $N$ has finite injective dimension.  However, in this case it seems to me that $N$ is a finitely generated projective $R$-module, so that the problem becomes trivial.  I am including the following argument for any case.

<B>Edit.</B> Something is wrong with the base case of the following induction.  I will try to fix it.

From the formulation of your question, it sounds like you are interested in associative but not-necessarily-commutative rings.  I cannot say anything about those.  However, for every commutative Noetherian ring $R$, for all maximal Cohen-Macaulay $R$-modules $M$ and $N$, the natural $R$-module homomorphism, $$\beta_{R,M,N}:\text{Hom}_R(M,R)\otimes_R N \to \text{Hom}_R(M,N),$$ is an isomorphism.  

First, since $\beta_{R,M,N}$ is a natural transformation of additive functors, this is obviously an isomorphism when $N$ is a finite free $R$-module.  Second, since this is a local problem, it suffices to prove the case when $R$ is a local, Noetherian, Gorenstein ring of some dimension $d$ (local Noetherian rings have finite Krull dimension).  The result is proved by induction on $d$.  I will use some results from Chapter 21 of the following. (<b>Aside:</b> This is the second time I have mentioned Chapter 21 on MathOverflow in two days.)

MR1322960 (97a:13001) <br>
Eisenbud, David(1-BRND) <br>
Commutative algebra. With a view toward algebraic geometry. <br>
Graduate Texts in Mathematics, 150. Springer-Verlag, New York, 1995. xvi+785 pp. <br> 
ISBN: 0-387-94268-8; 0-387-94269-6 <br>
13-01 (14A05) <br>

In the base case that $d=0$, by Proposition 21.11, both $M$ and $N$ are finite free $R$-modules.  Thus by the first observation above, $\beta_{R,M,N}$ is an isomorphism.  

Thus, by way of induction, assume that $d>0$, and assume that the result has been proved for smaller values of $d$.  Since $R$ is Gorenstein, it is Cohen-Macaulay, so that $\text{depth}_R(M) = d > 0$.  So there exists a system of parameters $(x_1,\dots,x_d)$ for $R$.  By Proposition 21.9, this system of parameters is simultaneously an $M$-sequence and an $N$-sequence.  Denote $\overline{R} = R/x_d R$, $\overline{M} = M/x_d M$, and $\overline{N} = N/x_d N$.
In particular, consider the $\overline{R}$-module homomorphism induced from $\beta_{R,M,N}$,
$$ \overline{\beta_{R,M,N}} : \text{Hom}_R(M,R)\otimes_R N \otimes_R \overline{R} \to \text{Hom}_R(M,N)\otimes_R \overline{R}.$$  By Proposition 21.13, to prove that $\beta_{R,M,N}$ is an isomorphism, it is equivalent to prove that $\overline{\beta_{R,M,N}}$ is an isomorphism.  

Since $R$ is a local Gorenstein ring of dimension $d$ and $x_d\in \mathfrak{m}_R$ is a nonzerodivisor, the quotient ring $\overline{R}$ is a local, Noetherian, Gorenstein ring (there are other references, but this does follow from Exercise 21.20).  Also, since $(x_1,\dots,x_d)$ is both an $M$-sequence and an $N$-sequence, by definition, also the induced system of parameters $(\overline{x}_1,\dots,\overline{x}_{d-1})$ is both an $\overline{M}$-sequence and an $\overline{N}$-sequence.  Therefore, by Proposition 21.9 again, both $\overline{M}$ and $\overline{N}$ are maximal Cohen-Macaulay $\overline{R}$-modules.  Thus, by the  induction hypothesis, the $\overline{R}$-module homomorphism,$$ \beta_{\overline{R},\overline{M},\overline{N}}: \text{Hom}_{\overline{R}}(\overline{M},\overline{R})\otimes_{\overline{R}} \overline{N} \to \text{Hom}_{\overline{R}}(\overline{M}, \overline{N}),  $$ is an isomorphism.  Finally, since $R$, $M$ and $N$ are maximal Cohen-Macaulay $R$-modules, by Proposition 21.12(b) (applied twice), the $\overline{R}$-module homomorphism $\beta_{\overline{R},\overline{M},\overline{N}}$ is equivalent to the $\overline{R}$-module homomorphism $\overline{\beta_{R,M,N}}$.  Therefore $\beta_{R,M,N}$ is an isomorphism.  By induction, the result holds for all $d$.