Given that, $\overrightarrow{a}=\frac{3}{2}\hat{k}, \overrightarrow{b}=\frac{2\hat{i}+2\hat{j}-\hat{k}}{2}$

$\Rightarrow \overrightarrow{a}+\overrightarrow{b}=\hat{i}+\hat{j}+\hat{k}$ and $\overrightarrow{a}-\overrightarrow{b}=-\hat{i}-\hat{j}+2\hat{k}$

$\Rightarrow \left|\overrightarrow{a}+\overrightarrow{b}\right|=\sqrt{3}$ and $\left|\overrightarrow{a}-\overrightarrow{b}\right|=\sqrt{6}$

Let the angle between them is $\theta$, then $\cos \theta =\frac{\left(\overrightarrow{a}+\overrightarrow{b}\right)·\left(\overrightarrow{a}-\overrightarrow{b}\right)}{\left|\overrightarrow{a}+\overrightarrow{b}\right|·\left|\overrightarrow{a}-\overrightarrow{b}\right|}$

$\Rightarrow \cos \theta =\frac{-1-1+2}{\sqrt{3}·\sqrt{6}}$

$\Rightarrow \cos \theta =0$

$\Rightarrow \theta =90°$