$$
\left\{\begin{array}{l}
\sqrt{x(1-y)}+\sqrt{y(1-x)}=\frac{1}{2}, \\
\sqrt{x(1-x)}+\sqrt{y(1-y)}=\frac{\sqrt{3}}{4} .
\end{array}\right.
$$

Solve the system of equations in the set of real numbers:
From (2), we know that $x(1-x) \geqslant 0, y(1-y) \geqslant 0$. Therefore, $0 \leqslant x \leqslant 1, 0 \leqslant y \leqslant 1$.
When $0 \leqslant x \leqslant 1, 0 \leqslant y \leqslant 1$, (1) also makes sense, so we can let
$$
x=\sin ^{2} \alpha, y=\sin ^{2} \beta,
$$

where $\alpha, \beta$ satisfy $0 \leqslant \alpha \leqslant \frac{\pi}{2}, 0 \leqslant \beta \leqslant \frac{\pi}{2}$.
Thus, the original system of equations becomes
$$
\left\{\begin{array}{l}
\sin \alpha \cdot \cos \beta+\sin \beta \cdot \cos \alpha=\frac{1}{2}, \\
\sin \alpha \cdot \cos \alpha+\sin \beta \cdot \cos \beta=\frac{\sqrt{3}}{4} .
\end{array}\right.
$$

Noting that $\sin \alpha \cdot \cos \alpha+\sin \beta \cdot \cos \beta=\frac{1}{2}(\sin 2 \alpha+ \sin 2 \beta)=\sin (\alpha+\beta) \cdot \cos (\alpha-\beta)$, we can further transform it into
$$
\left\{\begin{array}{l}
\sin (\alpha+\beta)=\frac{1}{2}, \\
\sin (\alpha+\beta) \cdot \cos (\alpha-\beta)=\frac{\sqrt{3}}{4} .
\end{array}\right.
$$
Dividing (4) by (3), we get
$\cos (\alpha-\beta)=\frac{\sqrt{3}}{2}$.
Since $0 \leqslant \alpha \leqslant \frac{\pi}{2}, 0 \leqslant \beta \leqslant \frac{\pi}{2}$, from (3) and (5) we have
$\alpha+\beta=\frac{\pi}{6}$ or $\frac{5 \pi}{6}, \alpha-\beta= \pm \frac{\pi}{6}$.

Therefore, the solutions to the original system of equations are
$$
\left(\frac{1}{4}, 0\right),\left(0, \frac{1}{4}\right),\left(1, \frac{3}{4}\right),\left(\frac{3}{4}, 1\right) \text {. }
$$