= Solution
With $X$ having rows $x_i^T=(1,g_i,m_i)$ and $\mu_i(\beta)=e^{x_i^T\beta}$, the <quasi-score equation> is
$$
X^T\{Y-\mu(\beta)\}=0.
$$
It is the same coefficient equation as for Poisson maximum likelihood, explaining why the two models have identical coefficient estimates.
Let $W=\operatorname{diag}(\widehat\mu_1,\ldots,\widehat\mu_n)$ and let coordinate $2$ denote gender. The model-based <standard error> is
$$
\operatorname{se}(\widehat\beta_1)
=\sqrt{\widehat\phi\,[(X^TWX)^{-1}]_{22}},
\qquad
\widehat\phi=\frac{X_P^2}{n-3}.
$$
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