Đáp án:
a) Để pt có 2 nghiệm phân biệt thì:
$\begin{array}{l}
\Delta > 0\\
\Rightarrow {\left( {m - 1} \right)^2} - 4\left( { - m - 4} \right) > 0\\
\Rightarrow {m^2} - 2m + 1 + 4m + 16 > 0\\
\Rightarrow {m^2} + 2m + 1 + 16 > 0\\
\Rightarrow {\left( {m + 1} \right)^2} + 16 > 0
\end{array}$
Vậy pt luôn có 2 nghiệm phân biệt với mọi m
$\begin{array}{l}
b)Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 1 - m\\
{x_1}{x_2} = - m - 4
\end{array} \right.\\
\left| {{x_1}} \right| = 3\left| {{x_2}} \right|\\
\Rightarrow \left[ \begin{array}{l}
{x_1} = 3{x_2}\\
{x_1} = - 3{x_2}
\end{array} \right.\\
+ Khi:{x_1} = 3{x_2}\\
\Rightarrow \left\{ \begin{array}{l}
4{x_2} = 1 - m\\
3x_2^2 = - m - 4
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{x_2} = \dfrac{{1 - m}}{4}\\
x_2^2 = \dfrac{{ - m - 4}}{3}
\end{array} \right.\\
\Rightarrow {\left( {\dfrac{{1 - m}}{4}} \right)^2} = \dfrac{{ - m - 4}}{3}\left( {dk:m < - 4} \right)\\
\Rightarrow 3\left( {{m^2} - 2m - 1} \right) = 16.\left( { - m - 4} \right)\\
\Rightarrow 3{m^2} - 6m - 3 = - 16m - 64\\
\Rightarrow 3{m^2} + 10m + 61 = 0\left( {vn} \right)\\
+ Khi:{x_1} = - 3{x_2}\\
\Rightarrow \left\{ \begin{array}{l}
- 2{x_2} = 1 - m\\
- 3x_2^2 = - m - 4
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{x_2} = \dfrac{{m - 1}}{2}\\
x_2^2 = \dfrac{{m + 4}}{3}\left( {m > - 4} \right)
\end{array} \right.\\
\Rightarrow {\left( {\dfrac{{m - 1}}{2}} \right)^2} = \dfrac{{m + 4}}{3}\\
\Rightarrow 3.\left( {{m^2} - 2m + 1} \right) = 4\left( {m + 4} \right)\\
\Rightarrow 3{m^2} - 6m + 3 = 4m + 16\\
\Rightarrow 3{m^2} - 10m - 13 = 0\\
\Rightarrow \left( {m + 1} \right)\left( {3m - 13} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
m = - 1\\
m = \dfrac{{13}}{3}
\end{array} \right.\left( {tmdk} \right)\\
Vậy\,m = - 1;m = \dfrac{{13}}{3}
\end{array}$
