The answer to question 2 is no. Take $A=\mathbb Z/8\mathbb Z$ where $\gamma$ acts by multiplication by $5$. Then $A^\Gamma= \ker( \cdot 4)= 2\mathbb Z/8\mathbb Z$ but $(\mathrm{id} +\gamma)(3)=6\cdot 3=2$, thus $H^2(\Gamma,A)$=0 (and obviously $A$ doesn't contain any induced module).

For the question 1 I do not really know what to say except that if $A$ is killed by 2, then $H^2(\Gamma,A)=0$ forces $A$ to be induced. However as you see in the example above, the "$H^2(\Gamma,A)=0$"-property does not pass to the 2-torsion subgroup $A[2]$ or the subquotients $A[2^i]/A[2^{i+1}]$, so I do not think this gives you anything when considering a general $A$. I also think that asking $H^2(\Gamma,A)=0$ doesn't really simplify the classification (so if we restrict to $2$-groups we just get finite $\mathbb Z_2[x]/(x^2-1)$-modules for which $\ker(x-1)=\mathrm{im}(1+x)$).

The answer to question 2 is yes. Take $A=\mathbb Z/8\mathbb Z$ where $\gamma$ acts by multiplication by $5$. Then $A^\Gamma= \ker( \cdot 4)= 2\mathbb Z/8\mathbb Z$ but $(\mathrm{id} +\gamma)(3)=6\cdot 3=2$, thus $H^2(\Gamma,A)$=0 (and obviously $A$ doesn't contain any induced module).

For the question 1 I do not really know what to say except that if $A$ is killed by 2, then $H^2(\Gamma,A)=0$ forces $A$ to be induced. However as you see in the example above, the "$H^2(\Gamma,A)=0$"-property does not pass to the 2-torsion subgroup $A[2]$ or the subquotients $A[2^i]/A[2^{i+1}]$, so I do not think this gives you anything when considering a general $A$. I also think that asking $H^2(\Gamma,A)=0$ doesn't really simplify the classification (so if we restrict to $2$-groups we just get finite $\mathbb Z_2[x]/(x^2-1)$-modules for which $\ker(x-1)=\mathrm{im}(1+x)$).