Đáp án:
$\begin{array}{l}
2)\\
m{x^2} + \left( {m - 1} \right)x + 2m + 1 \ge 0\forall x\\
\Rightarrow \left\{ \begin{array}{l}
m > 0\\
\Delta \le 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m > 0\\
{\left( {m - 1} \right)^2} - 4.m.\left( {2m + 1} \right) \le 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m > 0\\
{m^2} - 2m + 1 - 8{m^2} - 4m \le 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m > 0\\
7{m^2} + 6m - 1 \ge 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m > 0\\
\left( {7m - 1} \right)\left( {m + 1} \right) \ge 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m > 0\\
\left[ \begin{array}{l}
m \ge \frac{1}{7}\\
m \le - 1
\end{array} \right.
\end{array} \right.\\
Vậy\,m \ge \frac{1}{7}\\
3)\frac{\pi }{2} < a < \pi \\
\Rightarrow \cos a < 0\\
Do:{\sin ^2}a + {\cos ^2}a = 1\\
\Rightarrow \cos a = - \sqrt {1 - {{\sin }^2}a} = - \sqrt {1 - \frac{9}{{25}}} = - \frac{4}{5}\\
\Rightarrow \left\{ \begin{array}{l}
\tan a = \frac{{\sin a}}{{\cos a}} = \frac{3}{5}:\left( { - \frac{4}{5}} \right) = - \frac{3}{4}\\
\cot a = - \frac{4}{3}
\end{array} \right.
\end{array}$
