Solve: From $1-x \geqslant 0, x-\frac{1}{2} \geqslant 0 \Rightarrow \frac{1}{2} \leqslant x \leqslant 1$.
According to the problem, we have
$$\begin{aligned} y^{2} & =\left(\sqrt{1-x}+\sqrt{x-\frac{1}{2}}\right)^{2} \\ & =\frac{1}{2}+2 \sqrt{(1-x)\left(x-\frac{1}{2}\right)}: \\ & =\frac{1}{2}+2 \sqrt{-\left(x-\frac{3}{4}\right)^{2}+\frac{1}{16}} . \\ & \text { Let } t=-\left(x-\frac{3}{4}\right)^{2}+\frac{1}{16}(t \geqslant 0) . \end{aligned}$$

By the properties of quadratic functions and $\frac{1}{2} \leqslant x \leqslant 1$, we know that when $x=\frac{3}{4}$, $t_{\text {max }}=\frac{1}{16}, y_{\text {max }}=1$; when $x=\frac{1}{2}$ or $x=1$, $t_{\text {min }}=0, y_{\text {min }}=\frac{\sqrt{2}}{2}$. Therefore, $a=1, b=\frac{\sqrt{2}}{2} \Rightarrow a^{2}+b^{2}=\frac{3}{2}$.