Now we consider the case that $G$ has an almost square root $A$ with $x$ as above of two-power order greater than $1$ and with $x \in Z(G)$. We claim that $x$ does have order $2$.
We consider the eigenvalues of $A^{+}$ in the regular representation of $G$. The trivial representation of $G$ yields the eigenvalue $+\sqrt{|G|+1}$ with multiplicity $1$. Since $G^{+}$ acts as zero and $x$ acts as a scalar on every other irreducible representation, we conclude that if $x$ has order $2^{n} >1$, then every $2^{n}$-th root of unity other than $1$ occurs with mutiplicity $\frac{|G|}{2^{n}}$ as an eigenvalue of $G^{+}+x$ in the regular representation of $G$. Since $A^{+}$ is represented by an integer matrix in the regular representation of $G$, we conclude that algebraically conjugate eigenvalues occur with equal multiplicity, and that eigenvalues are closed under algebraic conjugation. Now $A^{+}$ must have some primitive $2^{n+1}$-th root of unity amongst its eigenvalues, and hence must have all $2^{n}$ primitive $2^{n+1}$-th roots of unity among its eigenvalues, with equal multiplicity $m$, say.
Consideration of the eigenvalues of $(A^{+})^{2}$ show that $m2^{n} = 2^{n-1}\frac{|G|}{2^{n}}.$ Thus $m = \frac{|G|}{2}.$ Hence the total multiplicity of primitive $2^{n+1}$-th roots of unity as eigenvalues of $A^{+}$ is $\frac{|G|}{2}.$
But if $2^{n} > 2$, a similar argument shows that the total multiplicity of primitive $2^{n}$-th roots of unity as eigenvalues of $A^{+}$ is also $\frac{|G|}{2}$, a contradiction, since $+\sqrt{|G|+1}$ is also an eigenvalue of $A^{+}.$ Thus when $x \neq 1$ has two-power order, $x$ must indeed have order $2$.
We may now also conclude that if a general finite group $G$ of even order has an almost square root $A$ with $(A^{+})^{2} = G^{+} + x$ where $x$ is a non-identity element of $2$-power order of $G$, then $x$ has order $2$ ( using the fact that $H = C_{G}(x)$ has a similar almost square root, and $x$ is now central).
Partial answer towards problem $5$: If $G$ is a finite group with an almost square root $A$, then a Sylow $2$-subgroup of $Z(G)$ has exponent dividing $4$.
For if $(A^{+})^{2} = G^{+} + x$, where (wlog) the order of $x$ is a power of $2$, then for every $2$-element $z \in Z(S)$, we know that $(zA^{+})^{2} = G^{+} + z^{2}x.$ Then by the above arguments, both $x$ and $z^{2}x$ have order dividing $2$. Hence $z^{4} = z^{4}x^{2} = 1$ and $z$ has order dividing $4$.