Given: ${\left(\frac{\sqrt{6}-\sqrt{2}}{4}+i\frac{\sqrt{6}+\sqrt{2}}{4}\right)}^{2020}$

$[\text{As we know} \sin \frac{5\mathrm{\pi }}{12}=\frac{\sqrt{3}+1}{2\sqrt{2}}=\frac{\sqrt{6}+\sqrt{2}}{4},$$\cos \frac{5\mathrm{\pi }}{12}=\frac{\sqrt{6}-\sqrt{2}}{4}]$

$={\left(\cos \frac{5\mathrm{\pi }}{12}+i\sin \frac{5\mathrm{\pi }}{12}\right)}^{2020}$

$\left[\text{As we know}, {\left(\cos \theta \pm \sin \theta \right)}^n=\cos n\theta \pm i\sin n\theta \right]$

$=\cos \left(\frac{2020\times 5\mathrm{\pi }}{12}\right)+i \sin \left(\frac{2020\times 5\mathrm{\pi }}{12}\right)$

$=\cos \left(\frac{505\times 5\mathrm{\pi }}{3}\right)+i \sin \left(\frac{505\times 5\mathrm{\pi }}{3}\right)$

$=\cos \left(842\mathrm{\pi }-\frac{\mathrm{\pi }}{3}\right)+i \sin \left(842\mathrm{\pi }-\frac{\mathrm{\pi }}{3}\right)$

$=\cos \frac{\mathrm{\pi }}{3}-i \sin \frac{\mathrm{\pi }}{3}$

$=\frac{1}{2}-\frac{\sqrt{3}}{2}i$