1)I don't understand how to use the universal coefficient theorem for group homology for to prove this.
The following reference explains the reasoning should be used, but I don't know how use a map $A\times A \to A$
This group is best understood in terms of the universal coefficient formula, i.e., in terms of the homology of the involved group. Hence, if $A$ is any abelian group we have $H_1(A)=A$ and the addition map $A\times A \to A $ induces a Pontryagin product on homology making $H^∗(A)$ a (graded) commutative algebra.
We also have that the square of an element of $H_1(A)$ is zero (in $H_2(A))$. This does not directly follow in the presence of $2$-torsion but it follows by functoriality; given $a\in H_1(A) = A$ we have a group homomorphism $\mathbb Z \to A$ taking 1 to a and we are reduced to showing that $1\cdot 1=0 \in H_2(Z)$ but $H_2(\mathbb Z)=0$.
Hence we get an algebra map $\bigwedge^∗A \to H^∗(A)$. This is an isomorphism in degrees $1$ and $2$ and an isomorphism in all degrees if $A$ is torsion free. This is proved easily by noting that both sides commutes with filtered direct limits so that one is reduced to the case when A is finitely generated and then using the Künneth formula to reduce to the case when $A$ is cyclic in which case $H_2(A)=0$ and $H_i(A)=0$ for $i\geq 2$, if $A=\mathbb Z$.
2) If $A$ is an abelian group, then $H_n(A,Q) = \bigwedge^n (A ⊗Q ) $?
