Let $S$ be the subgroup of $\operatorname{Gal}(M/H)$ generated by the inertia groups of the places $v_1,\cdots,v_s$ of $H$ ramifying in $\tilde{H}$. The index of the image of $S$ in $\operatorname{Gal}(\tilde{H}/H)$ is finite. Hence, replacing $H$ by a finite extension, we can assume that this image is isomorphic to $\mathbb Z_{p}^{d}$. Because $M/\tilde{H}$ is unramified, $\operatorname{Gal}(M/H)$ is a semi-direct product of $S$ with $X$.

Let $M_0$ be the maximal abelian extension of $H$ inside $M$ (note that $M_0$ contains $\tilde{H}$) and let $H_0$ be the maximal unramified extension of $H$ contained in $M_0$ (a finite extension).

By definition, $X/(t_1,\cdots,t_d)X$ is the maximal quotient of $X$ on which $\operatorname{Gal}(\tilde{H}/H)$ acts trivially. Because the action of $\operatorname{Gal}(\tilde{H}/H)$ on $X$ is by inner automorphism, this maximal quotient is also the quotient of $\operatorname{Gal}(M/H)$ by its derived subgroup and so the field $L$ fixed by $(t_1,\cdots,t_d)X$ is the maximal abelian extension of $H$ contained in $M$, and thus is equal to $M_0$. The answer to your question 2 is thus positive.

Moreover, as the places of $H$ ramified in $M_0$ are exactly the $v_i$ and as $M_0/\tilde{H}$ is unramified, we see that $S$ surjects on $\operatorname{Gal}(M_0/H_0)$ and hence that this $\mathbb Z_p$-modules has rank at most $d$. Hence, the rank of $\operatorname{Gal}(M_0/H)$ is less than $d$ and so the rank of $\operatorname{Gal}(M_0/\tilde{H})$ is zero. The module $X/(t_1,\cdots,t_d)$ is thus indeed a torsion $\mathbb Z_p$-module.

In fact, the answer above to your question 2) (resp. to your question 1)) exactly recapitulates the proof of the fact that $X$ is a noetherian $\Lambda$-module (resp. a torsion $\Lambda$-module). Good references include J-P.Serre *Classes des corps cyclotomiques (d'après K.Iwasawa)* (Séminaire Bourbaki, 1958) and K.Iwasawa *On $\mathbb Z_\ell$-extensions of Algebraic Number Fields* (Annals of Math.,1973).