Calculating $\mathrm{Var} (Z|Z|)$ Suppose $Z\sim N(0,1). $ How can I calculate $\mathrm{Var} (Z|Z|)$? 
I know $\mathrm{Var} (Z|Z|)= \mathrm{E}(Z^4)-\mathrm{E}^2(Z|Z|)$
 A: Your expression 
$$\text{Var}\big(Z \,\text{Abs}(Z)\big) = \boldsymbol{E}[Z^4] - \boldsymbol{E}[ Z \, \text{Abs}(Z) ]^2,$$
is correct, and a good starting point; here I am using $\text{Abs}(Z)$ in place of $|Z|$ so as to make the following lines with conditional expectations easier to read.  We calculate $\boldsymbol{E}[ Z \, \text{Abs}(Z) ]$,  by conditioning on whether $Z < 0$ or $Z > 0$. So
\begin{align}
\boldsymbol{E}[ Z \,\text{Abs}(Z) ] & = \boldsymbol{E}[ Z \,\text{Abs}(Z),\, | Z > 0 ]\boldsymbol{P}[ Z >0 ] +\boldsymbol{E}[ Z \,\text{Abs}(Z)\, | Z < 0 ]\boldsymbol{P}[ Z <0 ] \\
& = \frac12 \Big( \boldsymbol{E}[ Z \,\text{Abs}(Z)\, | Z > 0 ] +\boldsymbol{E}[ Z \,\text{Abs}(Z)\, | Z < 0 ] \Big) \\
& = \frac12 \Big(\boldsymbol{E}[ Z \,\text{Abs}(Z)\, | Z > 0 ] - \boldsymbol{E}[ Z \,\text{Abs}(Z)\, | Z >0 ] \Big) \\
& = 0.
 \end{align}
We used the symmetry of the normal distribution twice: once in claiming that $\boldsymbol{P}[ Z > 0] =\boldsymbol{P}[ Z < 0] = \frac12$, and then in the claim
$$\boldsymbol{E}[ Z \,\text{Abs}(Z)\, | Z < 0 ] = -\boldsymbol{E}[ Z \,\text{Abs}(Z)\, | Z > 0 ].$$
It follows then that the variance you ask for is
$$ \text{Var}\big(Z \text{Abs}(Z) \big) = \boldsymbol{E}[Z^4] = 3.$$
