(1) Let $f(x)=2^{x}, g(x)=4^{x}$, if $g[g(x)]>g[f(x)]>f[g(x)]$, find the range of $x$;
(2) If the range of the function $y=4^{x}-3 \cdot 2^{x}+3$ is $[1,7]$, find the maximum range of $x$.
(1) From the given information, we have
$$
\begin{array}{l}
g[g(x)]=4^{4^{x}}=4^{2^{2 x}}=2^{2^{2 x+1}}, \\
g[f(x)]=4^{2^{x}}=2^{2^{x+1}}, \\
f[g(x)]=2^{4^{x}}=2^{2^{2 x}},
\end{array}
$$

Since $g[g(x)]>g[f(x)]>f[g(x)]$, it follows that $2^{2^{2 x+1}}>2^{2^{x+1}}>2^{2^{2 x}}$, which simplifies to $2^{2 x+1}>2^{x+1}>2^{2 x}$, or $2 x+1>x+1>2 x$. Solving this, we get $0<x<1$.
(2) $y=4^{x}-3 \cdot 2^{x}+3=2^{2 x}-3 \cdot 2^{x}+3$. According to the problem, we have
$$
\left\{\begin{array}{l}
\left(2^{x}\right)^{2}-3 \cdot 2^{x}+3 \leqslant 7, \\
\left(2^{x}\right)^{2}-3 \cdot 2^{x}+3 \geqslant 1,
\end{array}\right.
$$

which simplifies to $\left\{\begin{array}{l}-1 \leqslant 2^{x} \leqslant 4, \\ 2^{x} \geqslant 2 \text { or } 2^{x} \leqslant 1,\end{array}\right.$ leading to $2 \leqslant 2^{x} \leqslant 4$ or $0<2^{x} \leqslant 1$. By the monotonicity of the function $y=2^{x}$, we get $x \in(-\infty, 0] \cup[1,2]$.