Đáp án:
$\begin{array}{l}
1)y = 3{\sin ^2}2x - 2{\cos ^3}2x\\
\Leftrightarrow y' = 3.2.2.\cos 2x.\sin 2x\\
- 2.3.2.\left( { - \sin 2x} \right).{\cos ^2}2x\\
\Leftrightarrow y' = 6.\sin 4x + 12\sin 2x.{\cos ^2}2x\\
= 6\sin 4x + 6.\sin 4x.\cos 2x\\
2)f\left( x \right) = \frac{{{x^3}}}{3} + m{x^2} + \left( {5m + 6} \right)x + 1\\
\Leftrightarrow f'\left( x \right) = {x^2} + 2mx + 5m + 6\\
f'\left( x \right) > 0\forall x\\
\Leftrightarrow {x^2} + 2mx + 5m + 6 > 0\forall x\\
\Leftrightarrow \Delta ' < 0\\
\Leftrightarrow {m^2} - 5m - 6 < 0\\
\Leftrightarrow \left( {m - 6} \right)\left( {m + 1} \right) < 0\\
\Leftrightarrow - 1 < m < 6\\
Vậy\, - 1 < m < 6
\end{array}$
