$Z$ ~ $N(0, 1)$. Find $\operatorname{Var}(Z)$ $Z$ ~ $N(0, 1)$. Find $\operatorname{Var}(Z)$
We know:
$E(Z) = 0$
$\operatorname{Var}(Z) = E(Z^2) - E(Z)^2 = E(Z^2) - 0^2 = E(Z^2)$
Thus, we need:
$E(Z^2) = \int_{-\infty}^{\infty} z^2 \frac{1}{\sqrt{2\pi}}e^{-1/2(z^2)}dz$
Using int by parts, $u = z, du = dz$, then $dv = z \frac{1}{\sqrt{2\pi}}e^{-1/2z^2}dz$, $v = -\frac{1}{\sqrt{2\pi}}e^{-1/2z^2}$
Then we have
$$\left[-\frac{z}{\sqrt{2\pi}}e^{-1/2z^2} \right]_{-\infty}^{\infty} + \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-1/2z^2}dz$$
I'm aware that $\left[-\frac{z}{\sqrt{2\pi}}e^{-1/2z^2} \right]_{-\infty}^{\infty} = (0-0) = 0$
No clue how to proceed. 

The standard method for evaluating $\int_{-\infty}^{\infty}e^{-\frac{1}{2}z^2}dz$ isn't an obvious one, but one worth remembering. It involves squaring it and turning it into a double integral.
First note that 
$$\int_{-\infty}^{\infty}e^{-\frac{1}{2}z^2}dz=2\int_{0}^{\infty}e^{-\frac{1}{2}z^2}dz$$
as $e^{-\frac{1}{2}z^2}$ is an even function.
Let $I:=\int_{0}^{\infty}e^{-\frac{1}{2}z^2}dz$. Also note that "$z$" is just a dummy variable, so $I=\int_{0}^{\infty}e^{-\frac{1}{2}x^2}dx=\int_{0}^{\infty}e^{-\frac{1}{2}y^2}dy$. 
$$
\begin{align}
I^2 &= \int_{0}^{\infty}e^{-\frac{1}{2}x^2}dx \int_{0}^{\infty}e^{-\frac{1}{2}y^2}dy
\\ &= \int_{0}^{\infty}\int_{0}^{\infty}e^{-\frac{1}{2}x^2}e^{-\frac{1}{2}y^2}dxdy
\\ &= \int_{0}^{\infty}\int_{0}^{\infty}e^{-\frac{1}{2}(x^2+y^2)}dxdy
\\ &= \int_{0}^{\frac{\pi}{2}}\int_{0}^{\infty}e^{-\frac{1}{2}r^2}rdrd\theta
\\ &= \frac{\pi}{2}\int_{0}^{\infty}re^{-\frac{1}{2}r^2}dr
\\ &= \frac{\pi}{2} \left[ -e^{-\frac{1}{2}r^2} \right]_{0}^{\infty}
\\ &= \frac{\pi}{2}
\end{align}
$$
... and hence
$$ I = \frac{\sqrt{2\pi}}{2} $$
which, together with your working, gives:
$$
\begin{align}
\mathrm{Var}(Z) &= E(Z^2)
\\ &= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-\frac{1}{2}z^2}dz
\\ &= \frac{2}{\sqrt{2\pi}} I
\\ &= 1
\end{align}
$$
as required!
