Đáp án:

$\begin{array}{l}
a)\left( { - 2;0} \right) \in \left( d \right)\\
 \Rightarrow 0 = \left( {m - 1} \right).\left( { - 2} \right) + 1\\
 \Rightarrow \left( {m - 1} \right).2 = 1\\
 \Rightarrow m - 1 = \dfrac{1}{2}\\
 \Rightarrow m = \dfrac{3}{2}\\
b)Xet:\dfrac{1}{2}{x^2} = \left( {m - 1} \right)x + 1\\
 \Rightarrow {x^2} - 2\left( {m - 1} \right)x - 2 = 0\\
 \Rightarrow \Delta ' = {\left( {m - 1} \right)^2} + 2 > 0
\end{array}$

=> pt hoành độ giao điểm luôn có 2 nghiệm pb

=> (d)  luôn cắt (P) tại 2 điểm phân biệt

c)

$\begin{array}{l}
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2\left( {m - 1} \right)\\
{x_1}{x_2} =  - 2
\end{array} \right.\\
A\left( {{x_1};\dfrac{1}{2}x_1^2} \right);B\left( {{x_2};\dfrac{1}{2}x_2^2} \right)\\
OA = \sqrt {x_1^2 + \dfrac{1}{4}x_1^4} ;OB = \sqrt {x_2^2 + \dfrac{1}{4}x_2^4} \\
 \Rightarrow {S_{OAB}} = \dfrac{1}{2}.OA.OB = 2\\
 \Rightarrow OA.OB = 4\\
 \Rightarrow O{A^2}.O{B^2} = 16\\
 \Rightarrow \left( {x_1^2 + \dfrac{1}{4}x_1^4} \right).\left( {x_2^2 + \dfrac{1}{4}x_2^4} \right) = 16\\
 \Rightarrow x_1^2.x_2^2 + \dfrac{1}{4}x_1^2x_2^2.\left( {x_1^2 + x_2^2} \right) + \dfrac{1}{{16}}x_1^4.x_2^4 = 16\\
 \Rightarrow {\left( { - 2} \right)^2} + \dfrac{1}{4}.{\left( { - 2} \right)^2}.\left( {x_1^2 + x_2^2} \right) + \dfrac{1}{{16}}.{\left( { - 2} \right)^4} = 16\\
 \Rightarrow \left( {x_1^2 + x_2^2} \right) = 11\\
 \Rightarrow {\left( {{x_1} + {x_2}} \right)^2} - 2{x_1}{x_2} = 11\\
 \Rightarrow 4{\left( {m - 1} \right)^2} - 2.\left( { - 2} \right) = 11\\
 \Rightarrow 4.{\left( {m - 1} \right)^2} = 7\\
 \Rightarrow {\left( {m - 1} \right)^2} = \dfrac{7}{4}\\
 \Rightarrow m = 1 \pm \dfrac{{\sqrt 7 }}{2}
\end{array}$