$Z$ ~ $N(0, 1)$. 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.
Using Fubini's theorem you can compute $(\int e^{-x^2})^2$=$\pi$ and in the same way that $\int e^{-\frac{x^2}{\alpha}}=\sqrt{\alpha\pi}$, which gives you the result you were looking for.
The answer is by definition of the law 1.

Moreover, you are trying to find the variance of a normal law knowing only the mean. Do you think it is even possible ? You could have any positive value for the variance.