Hochschild (co)homology and derived functors Suppose that $A$ is (complex) unital algebra. We will consider $A-A$ bimodules $M$: such a bimodule is the same as (say) left $A \otimes A^{op}$ module. Let us define $C_n(A,M)$ as $M \otimes A^{\otimes n}$ and $C^n(A,M)$ as $Hom(A^{\otimes n},M)$ ($C^0(A,M):=M$) and define maps $b$ and $\delta$ with the formulas $$b(m \otimes a_1 \otimes ... \otimes a_n)=ma_1 \otimes a_2 \otimes ... \otimes a_n+\sum_{j=1}^{n-1}(-1)^j m \otimes a_1 \otimes ... \otimes a_ja_{j+1} \otimes ... \otimes a_n+(-1)^n a_nm \otimes a_0 \otimes ... \otimes a_{n-1}$$ (on $C_0(A,M)=M$  we define $b=0$) 
$(\delta m)(a)=ma-am$ and $$(\delta f)(a_1,...,a_{n+1})=a_1f(a_2,...,a_{n+1}+\sum_{j=1}^{n}(-1)^jf(a_1,...,a_ja_{j+1},...,a_{n+1})+(-1)^{n+1}f(a_1,...,a_n)a_{n+1}$$
One checks that $b^2=0$ and $\delta^2=0$. Therefore we get (co)chain complexes $C^*(A,M)$ and $C_*(A,M)$. The underlying (co)homology is called Hochschild (co)homology. In order to prove that Hochschild (co)homology could be computed as the derived functor I have a problem with the following thing: according to J. Varilly, H. Figureoa, J.M Bracia-Bondia book one can identify $M \otimes_{A \otimes A^{op}} A^{\otimes (n+2)} \cong M \otimes A^{\otimes n}$ via the map: $m \otimes a_0 \otimes ... \otimes a_{n+1} \mapsto a_{n+1}ma_0 \otimes a_1 \otimes ... \otimes a_n$. 

Question 1: Why is this map an isomorphism? 

What is evident, is that this map is onto: but why it is one-to-one and why it is corerctly defined (autohrs didn't comment how $M \otimes_{A \otimes A^{op}} A^{\otimes (n+2)}$ and $M \otimes A^{\otimes n}$ are made into $A \otimes A^{op}$ modules: the problem is of course with $A \otimes A^{op}$ multiplication).
For cohomology authors claim that $Hom_{A \otimes A^{op}}(A^{\otimes (n+2)},M) \cong Hom(A^{\otimes n}, M)$ via the map $\alpha$ where $(\alpha T)(a_1,...,a_n):=T(1,a_1,...,a_n,1)$. Again, I'm not sure how $Hom_{A \otimes A^{op}}(A^{\otimes (n+2)},M)$ and $Hom(A^{\otimes n}, M)$ are made into $A \otimes A^{op}$ my second qestion is the following:

Question 2:  Why is this map an isomorphism? 

I always tried to be very careful checking all the properties which most of the authors finds trivial (I'm very careful especially when I meet something for the first time) so I would like to have a clear picture. Therefore I would be grateful for any help. 
 A: First, the $A\otimes A^{op}$-module structure on $A^{\otimes(n+2)}$ is of course by multiplication on the left on the first factor and multiplication on the right on the last. More precisely : $$(a\otimes b).(a_0\otimes\dots\otimes a_{n+2})=aa_0\otimes\dots a_{n+2}b$$
Hence, it is also isomorphic as an $A\otimes A^{op}$-module to $(A\otimes A^{op})\otimes A^{\otimes n}$ via
$$ a_0\otimes\dots\otimes a_{n+1}\mapsto (a_0\otimes a_{n+1})\otimes (a_1\otimes\dots\otimes a_n)$$
Now, $M$ is a right $A\otimes A^{op}$-module via
$$m.(a\otimes b)=b.m.a$$
I claim that for any ring $B$, any right-module $M$ and any left-module $N$
$$M\otimes_B (B\otimes N)=M\otimes N$$
via the isomorphism $m\otimes (b\otimes n)\mapsto mb \otimes n$ whose inverse is $m\otimes n\mapsto m\otimes (1\otimes n)$. With $B=A\otimes A^{op}$ and $N=A^{\otimes n}$, this is the isomorphism you seek. Explicitely, the inverse of the morphism you wrote is 
$$m\otimes a_1\otimes\dots\otimes a_n\mapsto m\otimes 1\otimes a_1\otimes\dots\otimes a_n\otimes 1$$
I let you check the dual statement, this is exactly the same module structures. Just note that we don't need any module structure on $M\otimes A^{\otimes n}$ nor do we need on $Hom(A^{\otimes n},M)$.
