8. The planar region $S=\left\{(x, y) \mid x, y \in\left[0, \frac{\pi}{2}\right], \sin ^{2} x-\sin x \sin y+\sin ^{2} y \leqslant \frac{3}{4}\right\}$ has an area equal to
8. $\frac{\pi^{2}}{6}$.

From
$$
\begin{aligned}
2\left(\sin ^{2} x-\sin x \sin y+\sin ^{2} y\right) & =1-\cos 2 x+\cos (x+y)-\cos (x-y)+1-\cos 2 y \\
& =2-2 \cos (x+y) \cos (x-y)+\cos (x+y)-\cos (x-y) \\
& =\frac{3}{2}-2\left(\cos (x+y)+\frac{1}{2}\right)\left(\cos (x-y)-\frac{1}{2}\right) \\
& \leqslant \frac{3}{2},
\end{aligned}
$$

we can get
$$
\left(\cos (x+y)+\frac{1}{2}\right)\left(\cos (x-y)-\frac{1}{2}\right) \geqslant 0,
$$

which means
$$
\left\{\begin{array} { l } 
{ x + y \leqslant \frac { 2 \pi } { 3 } , } \\
{ | x - y | \leqslant \frac { \pi } { 3 } }
\end{array} \text { or } \left\{\begin{array}{l}
x+y \geqslant \frac{2 \pi}{3}, \\
|x-y| \geqslant \frac{\pi}{3} .
\end{array}\right.\right.
$$

Combining $x, y \in\left[0, \frac{\pi}{2}\right]$, we can obtain the planar region as shown in Figure 3, whose area is $\left(\frac{\pi}{2}\right)^{2}-\left(\frac{\pi}{6}\right)^{2}-\frac{1}{2}\left(\frac{\pi}{3}\right)^{2}=\frac{\pi^{2}}{6}$.