This partial solution by Stuart Margolis: The
The answer to question 2$2$ is definitely yes, the number of elements required to generate G$G$ can't be more than D=2^(d-1)$D=2^{d-1}$. Suppose that G$G$ requires more than D$D$ generators, then every D$D$-tuple of elements of G$G$ must belong to a proper subgroup, which therefor must be solvable of derived length at most d-1$d-1$. So every D$D$-tuple of elements satisfies the identity (in D$D$ variables) that defines being solvable of derived length at most d-1$d-1$. So G$G$ itself must be solvable of derived length at most d-1$d-1$. This
This is just a special example of the fact that if a variety (in the Hanna Neumann sense) is defined by an n$n$-variable law then: a group which is not in the variety, but all its subgroups do belong to the variety, must be generated by at most n$n$ elements.