$\begin{array}{l}
y = 1 - 8{\sin ^2}x{\cos ^2}x + 2{\sin ^4}2x\\
= 1 - 2\left( {4{{\sin }^2}x{{\cos }^2}x} \right) + 2{\sin ^4}2x\\
= 1 - 2{\left( {2\sin x\cos x} \right)^2} + 2{\sin ^4}2x\\
= 1 - 2{\sin ^2}2x + 2{\sin ^4}2x\\
t = {\sin ^2}2x\left( {0 \le t \le 1} \right)\\
\Rightarrow y = f\left( t \right) = 2{t^2} - 2t + 1 = 2\left( {{t^2} - t + \dfrac{1}{4}} \right) + \dfrac{1}{2}\\
= 2{\left( {t - \dfrac{1}{2}} \right)^2} + \dfrac{1}{2} \ge \dfrac{1}{2}\\
\Rightarrow \min y = \dfrac{1}{2} \Rightarrow t = \dfrac{1}{2} \Rightarrow {\sin ^2}2x = \dfrac{1}{2}\\
\Leftrightarrow \dfrac{{1 - \cos 4x}}{2} = \dfrac{1}{2}\\
\Leftrightarrow \cos 4x = 0\\
\Leftrightarrow 4x = \dfrac{\pi }{2} + k\pi \\
\Leftrightarrow x = \dfrac{\pi }{8} + \dfrac{{k\pi }}{4}\left( {k \in \mathbb Z} \right)\\
+ f\left( 1 \right) = 1,f\left( 0 \right) = 1\\
\Rightarrow \max y = 1\\
\Rightarrow \left[ \begin{array}{l}
t = 1\\
t = 0
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
{\sin ^2}2x = 0\\
{\sin ^2}2x = 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sin 2x = 0\\
\sin 2x = 1\\
\sin 2x = - 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2x = k\pi \\
2x = \dfrac{\pi }{2} + k2\pi \\
2x = - \dfrac{\pi }{2} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{k\pi }}{2}\\
x = \dfrac{\pi }{4} + k\pi \\
x = - \dfrac{\pi }{4} + k\pi
\end{array} \right.\left( {k \in \mathbb Z} \right)
\end{array}$
