Modulus of normal random variable P and Q are variables such that-
$$P(P ≤ p,Q ≤ q)= \int_{-\infty}^{p} \int_{-\infty}^{q} \frac{1}{2\pi} e^{\left[ -\frac{1}{2}{(g^2 + h^2)} \right]}dgdh$$
Now random variables $V = mod(P)$ and $C = mod(Q)$. 
How do I prove that they are independent?
 A: You already have an answer for the independence part. I'll give an answer for the distribution part. You can easily see that $W_i\sim \mathcal{N}(0,1)$. I'll just write
 $W$. Let $x\geq 0$:
\begin{align} 
P(|W|\leq x)= P(-x \leq W\leq x)  = \Phi(x) - \Phi(-x) 
\end{align} 
Where $\Phi(\cdot) $ is the CDF of the standard normal. And surely $P(|W|\leq x) =0$ for $x<0$.
A: There is a standard result in probability theory that whenever $X$ and $Y$ are independent, then so are $f(X)$ and $g(Y)$ for (measurable) functions $f$ and $g$. This yields independence directly. 
For the distribution you need to apply the transformation formula for densities. This would be just some computational work.
A: The part about independence has been answered (see a comment and the answers).
As far as the joint distribution of the absolute values
$$P(Z_1 ≤ z_1,Z_2 ≤ z_2)= P(-z_1\leq W_1 ≤ z_1,-w_2\leq W_2 ≤ z_2)= $$
$$=\int_{-w_1}^{w_1} \int_{-w_2}^{w_2} \frac{1}{2\pi} e^{\left[ -\frac{1}{2}{(x^2 + y^2)} \right]}dxdy=\int_{-w_1}^{w_1}\frac{1}{2\pi} e^{ -\frac{1}{2}{x^2 } }\ dx\int_{-w_1}^{w_1}\frac{1}{2\pi} e^{ -\frac{1}{2}{y^2} }\ dy=$$
$$=(2\Phi(w_1)-1)(2\Phi(w_2)-1)$$
for positive $w_i$s -- otherwise zero. $\Phi$ is the distribution function of the standard normal case.
Note that the argumentation above  does not use independence directly. The latter simply follows from the shape of the  the original joint distribution.
