Both statements are true. To prove them, we will need the following lemma:
Lemma: Let $G$ be a finitely generated nilpotent group of class $k$ and let $H$ be a subgroup of $G$.
Then $\gamma_k(H)$ is a finite-index subgroup of $\gamma_k(G)$.
Proof: Iterated $k$-fold commutators induce a surjective homomorphism
$\phi\colon \wedge^k G^{\text{ab}} \rightarrow \gamma_k(G)$ of finitely generated abelian groups. Our hypotheses
imply that the image of $\wedge^k H^{\text{ab}}$ in $\wedge^k G^{\text{ab}}$ is finite-index, so the image of
the composition
$$\wedge^k H^{\text{ab}} \rightarrow \wedge^k G^{\text{ab}} \stackrel{\phi}{\rightarrow} \gamma_k(G)$$
has finite-index. But this implies that $\gamma_k(H)$ is finite-index in $\gamma_k(G)$, as desired. $\square$
Here's the proof of 1:
Theorem: Let $G$ be a finitely generated nilpotent group and let $H < G$ be a finite-index subgroup. Then
$H_n(H;\mathbb{Q}) \cong H_n(G;\mathbb{Q})$ for all $n$.
Proof: The proof will be by the degree $k$ of nilpotency of $G$. The base case $k=1$ means that $G$ is abelian,
and the theorem is easy. Assume now that the theorem is true for some $(k-1)$, and we will prove it for
$k$. Observe that $\gamma_k(H) < \gamma_k(G)$, and the lemma above says that
the abelian group $\gamma_k(H)$ is a finite-index subgroup of the finitely generated abelian group $\gamma_k(G)$. Setting
$\overline{H} = H / \gamma_k(H)$ and $\overline{G} = G/\gamma_k(G)$, we have a morphism between the central extensions
$$1 \rightarrow \gamma_k(H) \rightarrow H \rightarrow \overline{H} \rightarrow 1$$
and
$$1 \rightarrow \gamma_k(G) \rightarrow G \rightarrow \overline{G} \rightarrow 1.$$
We therefore get a morphism between the associated Hochschild-Serre spectral sequences. The map on
the $E^2$ page is of the form
$$H_p(\overline{H};H_q(\gamma_k(H);\mathbb{Q})) \rightarrow H_p(\overline{G};H_q(\gamma_k(G);\mathbb{Q})).$$
Since these are central extensions, this simplifies to a map
$$H_p(\overline{H};\mathbb{Q}) \otimes H_q(\gamma_k(H);\mathbb{Q}) \rightarrow H_p(\overline{G};\mathbb{Q}) \otimes H_q(\gamma_k(G);\mathbb{Q}).$$
By our inductive hypothesis, this is a tensor product of isomorphisms, so we conclude that
our two spectral sequences are the same and thus $H_n(H;\mathbb{Q}) \cong H_n(G;\mathbb{Q})$ for all $n$. $\square$
Here's the proof of 2:
Theorem: Let $G$ be a finitely generated nilpotent group and let $H < G$ be a finite-index subgroup. Then
the abelian groups $\gamma_n(H) / \gamma_{n+1}(H)$ and $\gamma_n(G) / \gamma_{n+1}(G)$ have the
same ranks for all $n$.
Proof: The proof will be by the degree $k$ of nilpotency of $G$. The base case $k=1$ means that $G$ is abelian,
and the theorem is easy. Assume now that the theorem is true for some $(k-1)$, and we will prove it for
$k$. Using our inductive hypothesis, it is enough to just deal with the bottom of the lower central
series and prove that the ranks of $\gamma_k(H)$ and $\gamma_k(G)$ are the same. But the lemma above
says that the abelian group $\gamma_k(H)$ is a finite-index
subgroup of $\gamma_k(G)$, so their ranks are definitely the same. $\square$
By the way, I can't help but point out a connection between your two questions. In his beautiful paper
Stallings, John,
Homology and central series of groups.
J. Algebra 2 (1965), 170-181.
Stallings proves the following theorem. Let $\gamma_k^{tf}$ be the torsion-free lower central series you define in your other question.
Theorem (Stallings): Let $f\colon H \rightarrow G$ be a group homomorphism inducing an isomorphism on $H_1(-;\mathbb{Q})$ and a surjection on $H_2(-;\mathbb{Q})$. Then $f$ induces isomorphisms between $(\gamma^{tf}_k(H)/\gamma^{tf}_{k+1}(H)) \otimes \mathbb{Q}$ and
$(\gamma^{tf}_k(G)/\gamma^{tf}_{k+1}(G)) \otimes \mathbb{Q}$ for all $k$. $\square$
This can be applied to the inclusion of a finite-index subgroup by the first theorem above, and together with the positive answer to your other question you can deduce a positive answer to your second one (but one that is obviously way over-powered!).
