= Solution
Write the <exponential dispersion family> in the convention
$$
f(z;\theta,\phi)=\exp\left\{\frac{z\theta-b(\theta)}{\phi}+c(z,\phi)\right\},\qquad \phi>0.
$$
Assume that the support does not depend on the parameters, that $\theta$ is in the interior of the <natural parameter space>, and that differentiation can pass under the normalizing integral or sum. Differentiating $\int f=1$ at fixed <dispersion parameter> gives
$$
0=\int\frac{z-b'(\theta)}{\phi}f(z;\theta,\phi)\,dz,
$$
so the <expectation> is $b'(\theta)$. A second differentiation gives
$$
0=\mathbb E\left[\frac{(Z-b'(\theta))^2}{\phi^2}-\frac{b''(\theta)}{\phi}\right].
$$
Consequently
$$
\boxed{\mathbb E Z=b'(\theta),\qquad \operatorname{Var}(Z)=\phi b''(\theta).}
$$
Here $b$ is the <cumulant function of an exponential family>; the same identities follow by differentiating $\log\mathbb E e^{tZ}=[b(\theta+\phi t)-b(\theta)]/\phi$. For a known observation weight $w$, replace $\phi$ by $\phi/w$. These regularity assumptions for <differentiation under the integral sign> matter: differentiation of a parameter-dependent support can introduce additional boundary terms.
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