= Solution
Choose \b[$g(t)=e^t$]. If $W$ has an <exponential distribution> with rate $r$, then for $c<r$,
$$
\mathbb E[e^{cW}]=\int_0^\infty r e^{-(r-c)t}\,dt=\frac r{r-c},
$$
while the <integral> diverges to $+\infty$ when $c\geq r$. Since $Z$ has rate $3$,
$$
\mathbb E[g(Z)]=\frac32,\qquad
\mathbb E[g(Z)^2]=3,\qquad
\boxed{\operatorname{Var}(g(Z))=\frac34.}
$$
For the other two rates, $\mathbb E[g(V)]=2/(2-1)=2$, whereas $\mathbb E[g(U)]=\int_0^\infty1\,dt=+\infty$. These computations establish exactly the required finite <variance>, finite <expected value>, and infinite <expected value>.
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