I think the answer is positive for the special case (2).
In this case every $(X,T)$-invariant measure $\mu$ is also $(\widetilde{X},T)$-invariant and hence admits an ergodic decompostion, say $\tau=\tau_\mu$. Since $\mu(X)=1$, $m(X)=1$ for $\tau$-a.e. $m\in E(\widetilde{X},T)$. In other words, the ergodic decomposition of $\mu$ is 'localized' on $(X,T)$. Note that $h_m(X,T)=h_m(\widetilde{X},T)$ for each $m$ with $m(X)=1$. Therefore
$h_\mu(X,T)=h_\mu(\widetilde{X},T)=\int_{E(T)}h_m(\widetilde{X},T)d\tau(m)=\int_{E(T)}h_m(X,T)d\tau(m)$.