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The Nielsen–Schreier theorem says that every subgroup of a free group is free. Realize $F_n$ as the fundamental group of the rose $R_n$, the graph with one vertex and $n$ oriented loops. A subgroup $H\le F_n$ of finite index of a subgroup $d$ corresponds to a connected $d$-sheeted covering graph $X\to R_n$. The graph $X$ has $d$ vertices and $dn$ unoriented edges. Choosing a spanning tree leavesedges outside the tree, and these freely generate ${\pi }_1(X)\cong H$. Thus the Nielsen–Schreier formula is

Take the free abelian group $G={\mathbb{Z}}^2$, whose rank of a group is two, and its finite-index subgroup $H=2\mathbb{Z}\times \mathbb{Z}$. The subgroup has index two and is again isomorphic to ${\mathbb{Z}}^2$, so its rank is two. The Nielsen–Schreier formula would instead give $1+2(2-1)=3$. Henceis a counterexample.

Yes. Take $G=F_2$. The free product satisfiesA connected three-sheeted cover of the two-petal rose has fundamental group of rank $1+3(2-1)=4$ by the Nielsen–Schreier formula. Thus $F_4$ is isomorphic to an index-three subgroup of $F_2$. A finite-index subgroup quasi-isometry then gives