Suppose that $R$ is a ring, that $M$, $N$ and $L$ are right $R$-modules, and that $N$ is an $R-R$-bimodule.


Is there any formula for $Ext^1 (M\otimes\_R N,L)$ or generally for $Ext^n (M\otimes\_R N,L)?$ 


Thanks in advance.



What is $Ext^1 (M\otimes_R N,L)$?
I assume from how the question is phrased that the OP is hoping for a kind of Hom-Tensor duality for Ext. The first comment below shows that in general


$$\operatorname{Ext}^n(M\otimes\_R N,L) \not \cong \operatorname{Ext}^n(M,\operatorname{Hom}\_R(N,L))$$


Instead, what you have are the well-known Ext-Tor relations. This makes sense: Ext is a left derived functor of Hom, so should play nicely with Tor, the right derived functor of $\otimes$.


In this answer, I'll show that Hom-Tensor duality of the sort that fails above for Ext holds for $\operatorname{Ext}\_{dw}$, the subgroup of Ext for chain complexes consisting of short exact sequences that are degreewise split. This $\operatorname{Ext}\_{dw}$ can be computed via the internal $\operatorname{Hom}$ in chain complexes. Let $S^n(M)$ be the chain complex with $M$ in degree $n$, 0 everywhere else, and all differentials 0. The Ext of chain complexes agrees with the Ext of modules, when they sit inside chain complexes via $S^0(-)$. Hence, $\operatorname{Ext}\_{dw}(S^0(A),S^0(B))$ is also a subgroup of the usual $\operatorname{Ext}(A,B)$.
First, note that:
$$\operatorname{Ext}\_{dw}^n(X,Y) \cong H\_n \operatorname{Hom}(X,Y) \cong \operatorname{Ch}(R)(X,\Sigma^{-n}Y)/\sim, $$
where $\sim$ is chain homotopy equivalence, see Lemma 2.1 of [this paper](http://www.ams.org/journals/tran/2004-356-08/S0002-9947-04-03416-6/S0002-9947-04-03416-6.pdf).


Let's apply this to your situation, letting $\otimes$ be the monoidal product in $\operatorname{Ch}(R)$: 
$$
\begin{aligned}
\operatorname{Ext}\_{dw}^n(S^0(M\otimes\_R N),S^0(L)) &\cong H\_n(\operatorname{Hom}(S^0(M\otimes\_R N),S^0(L))) \\
&\cong H\_n(\operatorname{Hom}(S^0(M)\otimes S^0(N),S^0(L))\\
&\cong H\_n(\operatorname{Hom}(S^0(M),\operatorname{Hom}(S^0(N),S^0(L)))) \\
&\cong \operatorname{Ext}\_{dw}^n(S^0(M),\operatorname{Hom}(S^0(N),S^0(L)))\\
&\cong \operatorname{Ext}\_{dw}^n(S^0(M),S^0(\operatorname{Hom}\_R(N,L)).
\end{aligned}
$$
Here I use that chain complex maps from $S^0(N)$ to $S^0(L)$ are the same as $R$-module maps from $N$ to $L$, since
$$\operatorname{Hom}(X,Y)\cong \prod\_i \operatorname{Hom}\_R(X\_i,Y\_{n+i}),$$
and for $S^0(N)$ and $S^0(L)$ the only place this product doesn't vanish is $\operatorname{Hom}\_R(N,L)$. We conclude that:


$$\operatorname{Ext}\_{dw}^n(M\otimes\_R N,L) \cong \operatorname{Ext}\_{dw}^n(M,\operatorname{Hom}\_R(N,L))$$


where modules sit inside chain complexes via the functor $S^0(-)$. 


