Example 2 Given that the domain of the function $y=f(x)$ is $[0,1]$, find: $(1) F(x)=$ $f\left(x^{2}\right)$; (2) $G(x)=f .(x+a)+f(x-a)$ the domain.

保留源文本的换行和格式，翻译结果如下：

Example 2 Given that the domain of the function $y=f(x)$ is $[0,1]$, find: $(1) F(x)=$ $f\left(x^{2}\right)$; (2) $G(x)=f .(x+a)+f(x-a)$ the domain.
Solution: (1) For $f\left(x^{2}\right)$ to be meaningful, it is necessary that the object on which $f$ acts, $x^{2}$, lies within the domain of $f(x)$, i.e., $0 \leqslant x^{2} \leqslant 1$. This yields $-1 \leqslant x \leqslant 1$. Therefore, the domain of $F(x)=f\left(x^{2}\right)$ is $[-1,1]$.
(2) According to the given conditions, the values of $x$ must satisfy
$\left\{\begin{array}{l}0 \leqslant x+a \leqslant 1 \\ 0 \leqslant x-a \leqslant 1\end{array}\right.$, which is $\left\{\begin{array}{l}-a \leqslant x \leqslant 1-a \\ a \leqslant x \leqslant 1+a\end{array}\right.$
When $a>\frac{1}{2}$, the domain is $\varnothing$.