Đáp án:

\(\begin{array}{l}
a)\left\{ \begin{array}{l}
m > 0\\
0 < m < \dfrac{4}{3}
\end{array} \right.\\
b)m < \dfrac{{ - 1 - \sqrt 5 }}{2}\\
B5:\\
\left[ \begin{array}{l}
m =  - 2\\
m = 1
\end{array} \right.
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)f\left( x \right) > 0\forall x\\
 \to \left\{ \begin{array}{l}
m > 0\\
9{m^2} - 4.3.m < 0
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
m > 0\\
9{m^2} - 12m < 0
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
m > 0\\
3m\left( {3m - 4} \right) < 0
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
m > 0\\
0 < m < \dfrac{4}{3}
\end{array} \right.\\
b)f\left( x \right) < 0\forall x\\
 \to \left\{ \begin{array}{l}
m + 1 < 0\\
1 - m\left( {m + 1} \right) < 0
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
m <  - 1\\
1 - {m^2} - m < 0
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
m <  - 1\\
\left[ \begin{array}{l}
m > \dfrac{{ - 1 + \sqrt 5 }}{2}\\
m < \dfrac{{ - 1 - \sqrt 5 }}{2}
\end{array} \right.
\end{array} \right.\\
 \to m < \dfrac{{ - 1 - \sqrt 5 }}{2}\\
B5:\\
f\left( x \right) = {m^2}x + mx + m - 2x + 2\\
 = \left( {{m^2} + m - 2} \right)x + m + 2\\
f\left( x \right) \ge 0\forall x\\
 \to \left( {{m^2} + m - 2} \right)x + m + 2 \ge 0\forall x\\
 \to \left( {m + 2} \right)\left( {m - 1} \right)x + m + 2 \ge 0\left( 1 \right)\\
Xét:m + 2 = 0\\
 \to m =  - 2\\
\left( 1 \right) \to 0x + 0 \ge 0\left( {ld} \right)\\
 \to m =  - 2\left( {TM} \right)\\
Xét:m - 1 = 0\\
 \to m = 1\\
\left( 1 \right) \to 0x + 3 \ge 0\left( {ld} \right)\\
KL:\left[ \begin{array}{l}
m =  - 2\\
m = 1
\end{array} \right.
\end{array}\)