Prove (1) $F(0)=c, F(-1)=a-b+c, F(1)=a-b+c$, so
$$F(x)=\frac{x(x-1)}{2} F(-1)-\left(x^{2}-1\right) F(0)+\frac{x(x+1)}{2} F(1),$$

Hence $2|F(x)|=|x(x-1)| \cdot|F(-1)|+2\left|x^{2}-1\right| \cdot|F(0)|$
$$\begin{aligned} & +|x(x+1)| \cdot|F(1)| \\ \leqslant & |x(x-1)|+2\left|x^{2}-1\right|+|x(x+1)| \end{aligned}$$

For $|x| \leqslant 1$, we have $0 \leqslant 1+x \leqslant 2,0 \leqslant 1-x \leqslant 2,0 \leqslant 1-x^{2} \leqslant 1$, thus
$$2|F(x)| \leqslant|x|(1-x)+2\left|x^{2}-1\right|+|x|(1+x)=2\left(|x|+1-|x|^{2}\right)$$

That is $|F(x)| \leqslant|x|+1-|x|^{2}=-\left(|x|-\frac{1}{2}\right)^{2}+\frac{5}{4} \leqslant \frac{5}{4}$.
(2) By the given conditions $|F(0)|=|c| \leqslant 1,|F(-1)|=|a-b+c| \leqslant 1$, $|F(1)|=|a+b+c| \leqslant 1$, since
$$c x^{2}+b x+a=c\left(x^{2}-1\right)+(a+b+c) \cdot \frac{1+x}{2}+(a-b+c) \cdot \frac{1-x}{2}$$

Therefore, for any $|x| \leqslant 1$ we have
$$\begin{aligned} \left|c x^{2}+b x+a\right|= & |c| \cdot\left|\left(x^{2}-1\right)\right|+|a+b+c| \cdot\left|\frac{1+x}{2}\right| \\ & \left.+|a-b+c| \cdot \left|\frac{1-x}{2}\right| \right. \\ \leqslant & \left|\left(x^{2}-1\right)\right|+\left|\frac{1+x}{2}\right|+\left|\frac{1-x}{2}\right| \\ \leqslant & \left(1-x^{2}\right)+\frac{1+x}{2}+\frac{1-x}{2} \\ = & 2-x^{2} \leqslant 2 \end{aligned}$$

That is $|G(x)| \leqslant 2$.