Known
The mean is:
$\displaystyle E(Y)=E(a(b+c))=\frac{1}{8}\int_0^{\infty}\int_0^{\ infty}\int_0^{\infty} a(b+c) e^{-a/2}e^{-b/2}e^{-c/2} ~da~db~dc $
which you should be able to do easily (using $\displaystyle \overline{a}=\overline{b}=\overline{c}=2$ )
Then You will need:
$\displaystyle E(Y^2)=E([a(b+c)]^2)=\frac{1}{8}\int_0^{\infty}\int_0^{\infty}\int_ 0^{\infty} [a(b+c)]^2 e^{-a/2}e^{-b/2}e^{-c/2} ~da~db~dc $
(use your knowlege of $\displaystyle \text{var}(a)$ , $\displaystyle \text{var}(b)$ and $\displaystyle \text{var}(c)$ as an aide to evaluating this, you should not need to do any actual integrals that way)
$\displaystyle \text{var}(Y)=E(Y^2)-[E(Y)]^2$
CB
If $\displaystyle X$,$\displaystyle Y$ and $\displaystyle Z$ are independedent RVs then their joint density:
$\displaystyle
p(x,y,z)=p(z)p(y)p(z)
$
where $\displaystyle p(x)$, $\displaystyle p(y)$ and $\displaystyle p(z)$ are the marginal densities of $\displaystyle X$, $\displaystyle Y$ and $\displaystyle Z$ respectivly.
CB