There are multiple formulations of an exponential family. But whichever one chooses to follow, the basic description is that if a random variable $X\sim p_{\theta}$ where $p_{\theta}$ is a probability model (pdf or pmf), then the family of distributions $P=\{p_{\theta}:\theta\in\Omega\}$ is a one-parameter exponential family (here $\theta$ is a scalar parameter) if $p_{\theta}$ can be expressed as
$$p_{\theta}(x)=\exp\{\eta(\theta)T(x)-B(\theta)\}h(x)\quad,\,x\in\mathscr{X}\,,\tag{*}$$
where $\mathscr{X}(\subseteq \mathbb R)$ is independent of $\theta$ and $\Omega$ is some (non-degenerate) subset of $\mathbb R$.
Here $h,T$ are functions of $x$ only and $\eta,B$ are functions of $\theta$ only.
The pdf $f(x\mid\theta)$ in the question is a Gumbel density with unit scale and (unknown) location $\theta$.
We have $$f(x\mid\theta)=\exp\{\eta(\theta)T(x)-B(\theta)\}h(x)\quad,\,x\in\mathbb R\quad,\theta\in\mathbb R,$$
where $\eta(\theta)=-e^{\theta},\,B(\theta)=-\theta,\,T(x)=e^{-x}$ and $h(x)=e^{-x}$.
So this is definitely a member of a one-parameter exponential family.
In fact if the scale parameter $\sigma$ (say) is known, then the general location-scale Gumbel pdf given by $$p(x)=\frac{1}{\sigma}e^{-(x-\theta)/\sigma}\exp\left(-e^{-(x-\theta)/\sigma}\right)\qquad,\,x\in\mathbb R\quad,\theta\in\mathbb R,\sigma>0$$
also belongs to the exponential family by the same logic.
If the scale $\sigma$ is unknown, then clearly $p(\cdot)$ no longer remains in the exponential family. This is because we cannot find a $T(x)$ and an $h(x)$ in the form $(*)$ which is free of $\sigma$.
First question
Inspecting the definition of the exponential family
$$
f_x(x;\theta) = c(\theta) g(x) e^{ \sum_{j=1}^l G_j(\theta) T_j(x) },
$$
one can say the following:
$T$ is a sufficient statistic. Condition on $T$, the conditional distribution is $g(x)$ (up to a normalization constant), which is independent of the parameter $\theta$. This is the definition of sufficiency. In fact, for the exponential family it is independent of $T$.
The term $e^{ \sum_{j=1}^l G_j(\theta) T_j(x) }$ determines the marginal distribution of $T$, via the choice of $G_j$'s.
$c(\theta)$ is a normalization constant so the density integrates to $1$.
Second question
As $G_j$'s are arbitrary, subject to measurability requirements etc., there is no general formula for computing moments. For the Poisson distribution, the first moment is simply
$$
e^{-\lambda} \sum_{k = 0}^{\infty} k \frac{\lambda^k}{k!}
= ( e^{-\lambda} \sum_{k = 1}^{\infty} \frac{\lambda^{k-1} }{(k-1)!}) \cdot \lambda = \lambda.
$$
The second moment is similar.
Best Answer
Hint: If $X$ were from exponential family, it would have finite expectation (you may even express it in terms of $A(\theta)$).