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

$h_\mu(X,T)=h_\mu(\widetilde{X},T)=\int_{E(T)}h_m(\widetilde{X},T)d\tau(m)=\int_{E(T)}h_m(X,T)d\tau(m)$.

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

$h_\mu(X,T)=h_\mu(\widetilde{X},T)=\int_{E(T)}h_m(\widetilde{X},T)d\tau(m)=\int_{E(T)}h_m(X,T)d\tau(m)$.

After André Caldas: I take a snapshot (link) of Theorem 6.4 in Chapter II of Mane's book:

alt text http://www.freeimagehosting.net/newuploads/e2770.jpg