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The answer to (1) is "yes." As Carnahan notes, you have $H\cap Z_G\subseteq Z_H$. Since $H/(Z_G\cap H)\cong HZ_G/Z_G$ is isomorphic to a subgroup of $G/Z_G$, then, $H/Z_H$ is a quotient of $H/(Z_G\cap H)$, hence isomorphic to a quotient of a subgroup of $G/Z_G$, so $\mathrm{rank}(H/Z_H) \leq \mathrm{rank}(G/Z_G)$.
The answer to (2) is "no". Take $G$ to be the relatively free group of class $2$ and rank $k$ (isomorphic to $F_k/(F_k)_3$, where $F_k$ is the free group of rank $k$ and $(F_k)_3$ is the third term of the lower central series of $G$). F_k$). Then$G^{\rm ab}\cong \mathbb{Z}^k$. Let$H=[G,G]$; then$H$is free abelian of rank$\binom{k}{2}$. Pick$k\gt 3$to get that the rank of$H/[H,H]$is greater than the rang of$G/[G,G]$. If you want to exclude the case where$H$is abelian itself, just add two of the original generators to the commutator subgroup so the abelianization is of rank$2+\binom{k-1}{2}$. For the modified version of (2) (see comments), it is still not true that$\mathrm{rank}(Z_H)\leq \mathrm{rank}(Z_G)$. Take$G$as above; then$Z_G=[G,G]$is free abelian of rank$\binom{k}{2}$. Take$H=\langle [G,G], x_1\rangle$, where$x_1$is one of the free generators of$G$. Then$H$is free abelian of rank$\binom{k}{2}+1$. 3 added 311 characters in body; added 9 characters in body The answer for to (1) is "yes." As Carnahan notes, you have$H\cap Z_G\subseteq Z_H$. Since$H/(Z_G\cap H)\cong HZ_G/Z_G$is isomorphic to a subgroup of$G/Z_G$, then,$H/Z_H$is a quotient of$H/(Z_G\cap H)$, hence isomorphic to a quotient of a subgroup of$G/Z_G$, so$\mathrm{rank}(H/Z_H) \leq \mathrm{rank}(G/Z_G)$. The answer to (2) is "no". Take$G$to be the relatively free group of class$2$and rank$k$(isomorphic to$F_k/(F_k)_3$, where$F_k$is the free group of rank$k$and$(F_k)_3$is the third term of the lower central series of$G$). Then$G^{\rm ab}\cong \mathbb{Z}^k$. Let$H=[G,G]$; then$H$is free abelian of rank$\binom{k}{2}$. Pick$k\gt 3$to get that the rank of$H/[H,H]$is greater than the rang of$G/[G,G]$. If you want to exclude the case where$H$is abelian itself, just add two of the original generators to the commutator subgroup so the abelianization is of rank$2+\binom{k-1}{2}$. For the modified version of (2) (see comments), it is still not true that$\mathrm{rank}(Z_H)\leq \mathrm{rank}(Z_G)$. Take$G$as above; then$Z_G=[G,G]$is free abelian of rank$\binom{k}{2}$. Take$H=\langle [G,G], x_1\rangle$, where$x_1$is one of the free generators of$G$. Then$H$is free abelian of rank$\binom{k}{2}+1$. 2 added 334 characters in body The answer for (2) is "no". Take$G$to be the relatively free group of class$2$and rank$k$(isomorphic to$F_k/(F_k)_3$, where$F_k$is the free group of rank$k$and$(F_k)_3$is the third term of the lower central series of$G$). Then$G^{\rm ab}\cong \mathbb{Z}^k$. Let$H=[G,G]$; then$H$is free abelian of rank$\binom{k}{2}$. Pick$k\gt 3$to get that the rank of$H/[H,H]$is greater than the rang of$G/[G,G]$. If you want to exclude the case where$H$is abelian itself, just add two of the original generators to the commutator subgroup so the abelianization is of rank$2+\binom{k-1}{2}$. For the modified version of (2) (see comments), it is still not true that$\mathrm{rank}(Z_H)\leq \mathrm{rank}(Z_G)$. Take$G$as above; then$Z_G=[G,G]$is free abelian of rank$\binom{k}{2}$. Take$H=\langle [G,G], x_1\rangle$, where$x_1$is one of the free generators of$G$. Then$H$is free abelian of rank$\binom{k}{2}+1\$.