Đáp án:

$\begin{array}{l}
1)f\left( x \right) > 0\forall x\\
 \Rightarrow \left( {m - 1} \right){x^2} - 2\left( {m + 1} \right).x + 3\left( {m - 2} \right) > 0\forall x\\
 \Rightarrow \left\{ \begin{array}{l}
m - 1 > 0\\
\Delta ' < 0
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
m > 1\\
{\left( {m + 1} \right)^2} - \left( {m - 1} \right).3.\left( {m - 2} \right) < 0
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
m > 1\\
{m^2} + 2m + 1 - 3{m^2} + 9m - 6 < 0
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
m > 1\\
2{m^2} - 11m + 5 > 0
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
m > 1\\
\left( {2m - 1} \right)\left( {m - 5} \right) > 0
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
m > 1\\
\left[ \begin{array}{l}
m < \frac{1}{2}\\
m > 5
\end{array} \right.
\end{array} \right.\\
 \Rightarrow m > 5\\
Vay\,m > 5\\
2)f\left( x \right) > 0\forall x\\
 \Rightarrow \left( {3 - m} \right).{x^2} - 2\left( {m + 3} \right).x + m + 2 > 0\forall x\\
 \Rightarrow \left\{ \begin{array}{l}
3 - m > 0\\
\Delta ' < 0
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
m < 3\\
{\left( {m + 3} \right)^2} - \left( {3 - m} \right)\left( {m + 2} \right) < 0
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
m < 3\\
{m^2} + 6m + 9 + {m^2} - m - 6 < 0
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
m < 3\\
2{m^2} + 5m + 3 < 0
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
m < 3\\
\left( {2m + 3} \right)\left( {m + 1} \right) < 0
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
m < 3\\
 - \frac{3}{2} < m <  - 1
\end{array} \right.\\
 \Rightarrow  - \frac{3}{2} < m <  - 1
\end{array}$