The first question about a distribution has no convenient general answer, because AFAIK nobody has assigned names to such distributions nor extensively studied and characterized them except when both $a$ and $b$ are $2$ or less.
Concerning the second question about the variances, as shorthand write $\mathbf{p}=(p_1,p_2,\ldots,p_n)$ and $\mathbf{x^p} = X_1^{p_1} X_2^{p_2} \cdots X_n^{p_n}$.
Recall that for the standard Normal distribution the $k^\text{th}$ moment is $0$ when $k$ is odd and otherwise equals $(k-1)!! = (k-1)(k-3)\cdots(3)(1)$. From this and the independence assumption, the expression for the expectation of $\mathbf{x^p}$ is immediate: it equals $0$ when one or more of the $p_i$ is odd and otherwise is the product of the $(p_i-1)!!$, which I will similarly abbreviate $(\mathbf{p-1})!!$.
By definition,
$$\text{Var}(\mathbf{x^p}) = \mathbb{E}[(\mathbf{x^p})^2] - \mathbb{E}[\mathbf{x^p}]^2 = (2\mathbf{p}\mathbf{-1})!! - ((\mathbf{p-1})!!)^2.$$
When the variables are scaled to have variance $\sigma^2$, $\mathbf{x^p}$ will be multiplied by $|\sigma|^{p_1+p_2+\cdots+p_n}$, whence its variance will be multiplied by $\sigma^{2(p_1+p_2+\cdots+p_n)}$.
Example
Let $\mathbf{p} = (2,4)$:
$(2\mathbf{p}\mathbf{-1})!! = 3!! 7!! = [3(1)][7(5)(3)(1)] = 315$;
$((\mathbf{p-1})!!)^2 = 1!!3!! = ([1][3(1)])^2 = 9$;
$\text{Var}(X_1^2 X_2^4) = (315 - 9)\sigma^{2(2+4)} = 306\sigma^{12}.$