$\let\res\restriction\def\N{\mathbb N}$By the discussion in comments, it is missing from the question that all the theories are assumed **consistent** (otherwise the answer is trivially yes, as everything is interpretable in the inconsistent extension) and **recursively axiomatizable** (otherwise the answer is trivially no, as there are $2^\omega$ pairwise incompatible complete extensions of ZFC, while only countably many of these are interpretable in any given consistent extension of ZC + rank).

With these caveats, the answer to both questions is **yes**, for general reasons that work for most other pairs of natural theories. In what follows, if $T$ is a theory with a fixed r.e. set of axioms, let $\Box_T$ denote the formalized provability predicate for $T$, and $T\res n$ the theory axiomatized by the axioms of $T$ with Gödel number below $n$.

**Proposition.** Let $S$ be an extension of $I\Delta_0+\mathrm{EXP}$, and $T$ be a recursively axiomatized theory such that
$$\tag{$*$}S\vdash\Box_{T\res n}\phi\implies T\vdash\phi$$
for all $n\in\N$ and all $T$-sentences $\phi$. Then every consistent r.e. extension of $T$ is interpretable in a consistent finite extension of $S$.

Observe that $(*)$ holds whenever $S\subseteq T$ (or just $S$ is $\Sigma_1$-conservative over $T$) and $T$ is a locally essentially reflexive theory, for example $S=\mathrm{ZC+rank}$ and $T=\mathrm{ZFC}$. But in fact, $(*)$ holds whenever $S$ is $\Sigma_1$-sound, which pretty much any natural theory is. Also, it is not difficult to show that if $S$ is locally essentially reflexive, then condition $(*)$ is *necessary*.

**Proof.** Let $U\supseteq T$ be recursively axiomatized and consistent. Then $(*)$ implies that
$$V=S+\{\mathrm{Con}_{U\res n}:n\in\N\}\equiv S+\{\neg\Box_{T\res n}\neg\phi:n\in\N,\phi\in U\}$$
is consistent; since it is an extension of $S$ by a r.e. set of $\Pi_1$-sentences, there exists a $\Pi_1$-sentence $\psi$ such that $S+\psi$ is a $\Sigma_1$-conservative extension of $V$ by a theorem of Lindström [1]. In particular, $S+\psi$ is consistent, and since it proves $\mathrm{Con}_{U\res n}$ for all $n\in\N$, it interprets $U$ by the usual interpretation existence lemma. QED

**Reference:**

[1] Per Lindström: *On partially conservative sentences and interpretability*, Proceedings of the American Mathematical Society 91 (1984), no. 3, pp. 436–443, doi: 10.2307/2045318.

definableotherwise the theory "ZFC + for all $n:2^{\aleph_n}=\aleph_{n+a_n}$" is not well defined, and so not effectively generated. So we only have countably many such extensions. $\endgroup$