$H^{ab}=\mathbb{Z}^n$, not trivial. Then you need to take the coinvariants $(H^{ab})_G=\mathbb{Z}^n/\operatorname{image}(A-I)$ Finally take product with $G^{ab}=\mathbb{Z}$, so the end result is $$ \mathbb{Z}\times\operatorname{coim}(A-I). $$
$H^{ab}=\mathbb{Z}^n$, not trivial. Then you need to take the coinvariants $(H^{ab})_G=\mathbb{Z}^n/\operatorname{image}(A-I)$ Finally take product with $G^{ab}=\mathbb{Z}$, so the end result is $$ \mathbb{Z}\times\operatorname{coim}(A-I). $$