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Work at a regular admissible parameter value with $\varphi \ne 0$, so the score identities in the question apply. For one observation from the exponential dispersion family, the score function and second derivative areThe zero expected score immediately gives $\mathbb{E}Y=b^{\prime }(\theta )$. The equality between expected negative curvature and score variance then givesThusThe admissible density parameters necessarily make this variance nonnegative; the algebra does not require an unstated sign assumption beyond the regular family and $\varphi \ne 0$.

With independent observations and canonical parameters ${\theta }_i=x_i^T\beta$, the chain rule gives the vector score functionA finite interior maximum-likelihood estimator satisfies its score equation, provingDifferentiating again, using ${\nabla }_{\beta }{\mu }_i=b^{\prime \prime }(x_i^T\beta )x_i$, givesThis Hessian is independent of the observations, so taking expectation gives exactly the requested Fisher information matrix. In particular, $V$ is not the inverse observation-variance matrix: the linear predictor here is the canonical parameter, not the mean.

For the two-group Poisson regression, $\varphi =1$, $b(\theta )=e^{\theta }$, and ${\mu }_i=e^{x_i^T\beta }$. Put $a=e^{\alpha }$, $b=e^{\alpha +\nu }$, using $b$ here for the second group's mean. There are $2n$ observations, andHence the total Fisher information matrix and its inverse areThe theorem on asymptotic normality of a maximum likelihood estimator says that in a regular identifiable model with nonsingular information the estimator is approximately multivariate normal, centred at the true parameter and with covariance equal to the inverse total information. Here one can regard one observation from each group as an independent pair and apply its usual iid version to $n$ such pairs. ThereforeFor a direct check, the second score equation gives the fitted second-group mean ${\overline{Y}}_B$, and the first then gives the fitted first-group mean ${\overline{Y}}_A$. Thus the Poisson log-rate ratio estimate is $\hat{\nu }=\log {\overline{Y}}_B-\log {\overline{Y}}_A$. The delta method, using independent sample-mean variances $a/n$ and $b/n$, gives the same variance $(a^{-1}+b^{-1})/n$. If a group total is zero, its finite log-mean estimator fails to exist; with fixed positive true means this event has probability tending to zero exponentially as $n$ grows.