Đáp án:
$\begin{array}{l}
a)A\left( { - 1;5} \right) \in y = 3x + {m^2} - 1\\
\Leftrightarrow 5 = 3.\left( { - 1} \right) + {m^2} - 1\\
\Leftrightarrow {m^2} = 9\\
\Leftrightarrow m = 3;m = - 3\\
Vậy\,m = 3;m = - 3\\
b)Xet:\left( P \right):y = {x^2}\\
\left( d \right):y = 3x + {m^2} - 1\\
\Leftrightarrow {x^2} = 3x + {m^2} - 1\\
\Leftrightarrow {x^2} - 3x - {m^2} + 1 = 0\\
\Delta = 9 - 4\left( { - {m^2} + 1} \right)\\
= 4{m^2} + 5 > 0
\end{array}$
=> chúng luôn cắt nhau tại 2 điểm phân biệt
$\begin{array}{l}
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 3 > 0\\
{x_1}{x_2} = - {m^2} + 1
\end{array} \right.\\
+ TH1:Khi:{x_1} \le 0 < {x_2}\\
\Leftrightarrow \left| {{x_1}} \right| + 2\left| {{x_2}} \right| = 3\\
\Leftrightarrow - {x_1} + 2{x_2} = 3\\
\Leftrightarrow - {x_1} + 2{x_2} = {x_1} + {x_2} = 3\\
\Leftrightarrow \left\{ \begin{array}{l}
3{x_2} = 6\\
{x_1} = 3 - {x_2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{x_2} = 2\\
{x_1} = 1\left( {ktm} \right)
\end{array} \right.\\
+ Khi:{x_2} \le 0 < {x_1}\\
\Leftrightarrow \left| {{x_1}} \right| + 2\left| {{x_2}} \right| = 3\\
\Leftrightarrow {x_1} + 2\left( { - {x_2}} \right) = 3\\
\Leftrightarrow \left\{ \begin{array}{l}
{x_1} - 2{x_2} = 3\\
{x_1} + {x_2} = 3
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{x_2} = 0\\
{x_1} = 3
\end{array} \right.\\
\Leftrightarrow 0.3 = 0 = - {m^2} + 1\\
\Leftrightarrow {m^2} = 1\\
\Leftrightarrow m = 1;m = - 1\\
+ Khi:0 < {x_1} < {x_2}\\
\Leftrightarrow \left| {{x_1}} \right| + 2\left| {{x_2}} \right| = 3\\
\Leftrightarrow \left\{ \begin{array}{l}
{x_1} + 2{x_2} = 3\\
{x_1} + {x_2} = 3
\end{array} \right.\\
\Leftrightarrow {x_2} = 0\left( {ktm} \right)
\end{array}$
Vậy $m = 1;m = - 1$
