Q1 is pretty basic.

$\mu_Z=E[Z]=E\left[\dfrac{X-\mu_X}{\sigma_X}\right]=$

$\dfrac{E[X]-\mu_X}{\sigma_X}=$

$\dfrac{\mu_X - \mu_X}{\sigma_X}=0$

$\mu_Z=0$

$var[Z]=E[\left(Z-\mu_Z\right)^2]=$

$var[Z]=E[Z^2]=$

$E\left[\left(\dfrac{X-\mu_X}{\sigma_X}\right)^2\right]=$

$\dfrac{1}{\sigma_X^2}E\left[\left(X-\mu_X\right)^2\right]=$

$\dfrac{1}{\sigma_X^2}var[X]=$

$\dfrac{1}{\sigma_X^2}\sigma_X^2=1$

$\sigma_Z=\sqrt{var[Z]}=\sqrt{1}=1$

Q2

$\large A\displaystyle{\int_{-\frac n 2}^{\frac n 2}}\cos(x)~dx=1$

you can do the integration and solve for A

$\large \mu_G=E[G]=A\displaystyle{\int_{-\frac n 2}^{\frac n 2}}x\cos(x)~dx$

$\large var[G]=A\displaystyle {\int_{-\frac n 2}^{\frac n 2}}(x-\mu_G)^2\cos(x)~dx$

$\sigma_G=\sqrt{var[G]}$