Đáp án:

$\begin{array}{l}
1)\\
a)A,B \in \left( d \right)\\
 \Rightarrow \left\{ \begin{array}{l}
 - 2 = \left( {k - 1} \right).0 + n\\
0 = \left( {k - 1} \right).\left( { - 1} \right) + n
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
n =  - 2\\
 - k + 1 + n = 0
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
n =  - 2\\
k = n + 1 =  - 1
\end{array} \right.\\
Vậy\,k =  - 1;n =  - 2\\
b)\left( d \right)//\left( \Delta  \right)\\
 \Rightarrow \left\{ \begin{array}{l}
k - 1 = 1\\
n \ne 2 - k
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
k = 2\\
n \ne 0
\end{array} \right.\\
Vậy\,k = 2;n \ne 0\\
2)\\
n = 2 \Rightarrow \left( d \right):y = \left( {k - 1} \right).x + 2\\
C\left( {x;0} \right);B \in Ox\\
A \in Oy\\
 \Rightarrow \left\{ \begin{array}{l}
OA = 2\\
OB = 1\\
OC = \left| x \right|
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
{S_{OAC}} = \dfrac{1}{2}OA.OC = \dfrac{1}{2}.2.\left| x \right| = \left| x \right|\\
{S_{OAB}} = \dfrac{1}{2}.OA.OB = \dfrac{1}{2}.2.1 = 1
\end{array} \right.\\
{S_{OAC}} = 2{S_{OAB}}\\
 \Rightarrow \left| x \right| = 2.1 = 2\\
 \Rightarrow \left[ \begin{array}{l}
x = 2\\
x =  - 2
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
C\left( {2;0} \right) \in \left( d \right)\\
C\left( { - 2;0} \right) \in \left( d \right)
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
0 = \left( {k - 1} \right).2 + 2\\
0 = \left( {k - 1} \right).\left( { - 2} \right) + 2
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
k = 0\\
k = 2
\end{array} \right.\\
Vậy\,k = 0\,hoặc\,k = 2
\end{array}$