8. Planar Region
$$
\begin{aligned}
S= & \left\{(x, y) \left\lvert\, x 、 y \in\left[0, \frac{\pi}{2}\right]\right.,\right. \\
& \left.\sin ^{2} x+\sin x \cdot \sin y+\sin ^{2} y \leqslant \frac{3}{4}\right\}
\end{aligned}
$$

The area of the region is
8. $\frac{\pi^{2}}{6}$.
$$
\begin{aligned}
\text { Given } & 2\left(\sin ^{2} x-\sin x \cdot \sin y+\sin ^{2} y\right) \\
= & 2-2 \cos (x+y) \cdot \cos (x-y)+ \\
& \cos (x+y)-\cos (x-y) \\
= & \frac{3}{2}-2\left(\cos (x+y)+\frac{1}{2}\right)\left(\cos (x-y)-\frac{1}{2}\right) \\
\leqslant & \frac{3}{2} \\
\Rightarrow & \left(\cos (x+y)+\frac{1}{2}\right)\left(\cos (x-y)-\frac{1}{2}\right) \geqslant 0 \\
\Rightarrow & \left\{\begin{array} { l } 
{ x + y \leqslant \frac { 2 \pi } { 3 } , } \\
{ | x - y | \leqslant \frac { \pi } { 3 } }
\end{array} \text { or } \left\{\begin{array}{l}
x+y \geqslant \frac{2 \pi}{3}, \\
|x-y| \geqslant \frac{\pi}{3} .
\end{array}\right.\right.
\end{aligned}
$$

Combining $x, y \in\left[0, \frac{\pi}{2}\right]$, the area of the required plane region is $\frac{\pi^{2}}{6}$.