(Edited following Emil Jerabek's coment below) From results of L. Pyber (and implicitly, C. Sims) it appears likely that $\frac{f(n)}{g(n)} \to 1$ as $n \to \infty,$ where $f(n)$ is the number of isomorphism types of nilpotent groups of order $n$ and $g(n)$ is the number of isomorphism types of all groups of order $n,$ so minor modifications should yield the same answer for question 2 (which is a cumulative version- note also that all nilpotent groups are solvable).
Also, the asymptotic behaviour of the number of isomorphism types of Abelian groups of order $n$ and the number of isomorphism types of nilpotent groups of order $n$ are known: both are multiplicative, so it suffices to consider the case of $p$-groups. The number of isomorphism types of Abelian groups of order $p^{k}$ is $p(k),$ the number of partitions of $k,$ which behaves like $e^{c \sqrt{k}}$ for some (known!) constant $c.$ The number of isomorphism types of groups of order $p^{k}$ is asymptotically around $p^{\frac{2k^{3}}{27}}$ (proved by C. Sims and G. Higman). This suggests that the limit of question 1 should be zero, though again you ask for a cumulative version.