\[\begin{array}{l}
1)\,\,\,{\sin ^8}x + {\cos ^8}x = \frac{1}{8}\\
\Leftrightarrow \left( {{{\sin }^2}x + {{\cos }^2}x} \right)\left( {{{\sin }^4}x - {{\sin }^2}x{{\cos }^2}x + {{\cos }^4}x} \right) = \frac{1}{8}\\
\Leftrightarrow {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^2} - 3{\sin ^2}x{\cos ^2}x = \frac{1}{8}\\
\Leftrightarrow 1 - \frac{3}{4}{\sin ^2}2x = \frac{1}{8}\\
\Leftrightarrow \frac{3}{4}{\sin ^2}2x = \frac{7}{8}\\
\Leftrightarrow {\sin ^2}2x = \frac{7}{6}\,\, > 1\,\\
\Rightarrow pt\,\,\,VN.
\end{array}\]
\[\begin{array}{l}
2)\,\,{\cos ^2}x + {\cos ^2}2x + {\cos ^2}3x + {\cos ^2}4x = \frac{3}{2}\\
\Leftrightarrow 2{\cos ^2}x + 2{\cos ^2}2x + 2{\cos ^2}3x + 2{\cos ^2}4x = 3\\
\Leftrightarrow 2{\cos ^2}x - 1 + 2{\cos ^2}2x - 1 + 2{\cos ^2}3x - 1 + 2{\cos ^2}4x = 0\\
\Leftrightarrow \cos 2x + \cos 4x + \cos 6x + 2{\cos ^2}4x = 0\\
\Leftrightarrow \cos 2x + \cos 6x + \cos 4x + 2{\cos ^2}4x = 0\\
\Leftrightarrow 2\cos 4x.\cos 2x + \cos 4x\left( {1 + 2\cos 4x} \right) = 0\\
\Leftrightarrow \cos 4x\left( {2\cos 2x + 1 + 2\cos 4x} \right) = 0\\
\Leftrightarrow \cos 4x\left( {4{{\cos }^2}2x + 2\cos 2x} \right) = 0\\
\Leftrightarrow 2\cos 4x.\cos 2x\left( {2\cos 2x + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos 4x = 0\\
\cos 2x = 0\\
\cos 2x = - \frac{1}{2}
\end{array} \right..
\end{array}\]
