$\begin{array}{l}
a){\sin ^4}2x + {\cos ^4}2x = \sin 2x\cos 2x\\
\Leftrightarrow {\left( {{{\sin }^2}2x + {{\cos }^2}2x} \right)^2} - 2{\sin ^2}2x{\cos ^2}2x = \sin 2x\cos 2x\\
\Leftrightarrow 1 - 2{\left( {\sin 2x\cos 2x} \right)^2} - \sin 2x\cos 2x = 0\\
\Leftrightarrow 2{\left( {\sin 2x\cos 2x} \right)^2} + \sin 2x\cos 2x - 1 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin 2x\cos 2x = -1\\
\sin 2x\cos 2x = \dfrac{1}{2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sin 4x = -2(L)\\
\sin 4x = 1
\end{array} \right.\\
\Rightarrow \sin 4x = 1\\
\Leftrightarrow 4x = \dfrac{{ \pi }}{2} + k2\pi \\
\Leftrightarrow x = \dfrac{\pi }{8} + \dfrac{{k\pi }}{2}\left( {k \in \mathbb{Z}} \right)\\
b)\cos 2x + {\sin ^2}x + 2\cos x + 1 = 0\\
\Leftrightarrow 2{\cos ^2}x + 2\cos x + {\sin ^2}x = 0\\
\Leftrightarrow 2{\cos ^2}x + 2\cos x + 1 - {\cos ^2}x = 0\\
\Leftrightarrow {\cos ^2}x + 2\cos x + 1 = 0\\
\Leftrightarrow {\left( {\cos x + 1} \right)^2} = 0\\
\Leftrightarrow \cos x = - 1\\
\Leftrightarrow x = \pi + k2\pi \left( {k \in \mathbb{Z}} \right)
\end{array}$
