$\begin{array}{l}
4\left( {{{\sin }^4}x + {{\cos }^4}x} \right) - 8\left( {{{\sin }^6}x + {{\cos }^6}x} \right) - 4{\sin ^2}4x = m\\
\Leftrightarrow 4\left[ {{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}^2} - 2{{\sin }^2}x{{\cos }^2}x} \right] - 8\left[ {{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}^3} - 3{{\sin }^2}x{{\cos }^2}x\left( {{{\sin }^2}x + {{\cos }^2}x} \right)} \right] - 4{\sin ^2}4x = m\\
\Leftrightarrow 4\left( {1 - \frac{1}{2}{{\sin }^2}2x} \right) - 8\left( {1 - \frac{3}{4}{{\sin }^2}2x} \right) - 4{\sin ^2}4x = m\\
\Leftrightarrow 4 - 2{\sin ^2}2x - 8 + 6{\sin ^2}2x - 4{\sin ^2}4x = m\\
\Leftrightarrow - 4 + 4{\sin ^2}2x - 4{\sin ^2}4x = m\\
\Leftrightarrow - 4{\cos ^2}2x - 4\left( {1 - {{\cos }^2}4x} \right) = m\\
\Leftrightarrow - 2\left( {1 + \cos 4x} \right) - 4 + 4{\cos ^2}4x = m\\
\Leftrightarrow 4{\cos ^2}4x - 2\cos 4x - 6 = m\\
Dat\,\cos 4x = t \in \left[ { - 1;1} \right] \Rightarrow 4{t^2} - 2t - 6 = m\\
Xet\,ham\,f\left( t \right) = 4{t^2} - 2t - 6\,la\,ham\,so\,bac\,hai\,co\,hoanh\,do\,dinh\,t = \frac{1}{4}\\
Tu\,bbt\,suy\,ra\,de\,pt\,vo\,nghiem\,thi\,\left[ \begin{array}{l}
m > 0\\
m < - \frac{{25}}{4}
\end{array} \right.
\end{array}$
