= Solution
For $\lambda>0$, write the <Poisson distribution> mass function in <exponential dispersion family> form:
$$
f(y;\lambda)=\exp\left\{\frac{y\theta-b(\theta)}{a(\phi)}+c(y,\phi)\right\},\quad
\theta=\log\lambda,\quad b(\theta)=e^\theta,\quad a(\phi)=\phi=1,\quad c(y,1)=-\log(y!).
$$
Thus the <natural parameter of an exponential family> is \b[$\theta=\log\lambda$] and the <dispersion parameter> is \b[fixed at $1$]. The counting-measure support is $0,1,\ldots$; there is no independent free dispersion parameter in an ordinary Poisson model. The degenerate boundary $\lambda=0$ is obtained as a limit, rather than a finite natural parameter.
The <exponential dispersion family> identities give $\mathbb EY=b'(\theta)$ and $\operatorname{Var}(Y)=a(\phi)b''(\theta)$. Since both derivatives of $e^\theta$ equal $e^\theta$,
$$
\boxed{\mathbb EY=\lambda,\qquad\operatorname{Var}(Y)=\lambda.}
$$
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