Solution:

$\cos 570^{\circ}=\cos \left(6 \times 90^{\circ}+30^{\circ}\right)$

$\cos 570^{\circ}=-\cos 30^{\circ}=-\frac{\sqrt{3}}{2}$

\begin{aligned} \sin 510^{\circ} &=\sin \left(6 \times 90^{\circ}-30^{\circ}\right) \ &=\sin 30^{\circ}=\frac{1}{2} \end{aligned}

\begin{aligned} \sin \left(-330^{\circ}\right) &=-\sin \left(330^{\circ}\right) \ &=-\sin \left(4 \times 90^{\circ}-30^{\circ}\right)=-\left(-\sin 30^{\circ}\right)=\frac{1}{2} \ \cos \left(-390^{\circ}\right) &=\cos 390^{\circ} \ &=\cos \left(4 \times 90^{\circ}+30^{\circ}\right)=\cos 30^{\circ}=\frac{\sqrt{3}}{2} \end{aligned}

$\therefore \cos 570^{\circ} \sin 510^{\circ}+\sin \left(-330^{\circ}\right) \cos \left(-390^{\circ}\right)$

$\quad=\left(-\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right)$

$\quad=0$