Đáp án:

\(\begin{array}{l}
a)\,\,\,m =  \pm 1\\
b)\,\,\,m = 4 \pm 2\sqrt 3 .
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
d:\,\,\,y = mx - m + 2\,\,\,\,\left( {m \ne 0} \right)\\
d\,\,\,\,cat\,\,\,cac\,\,\,truc\,\,\,toa\,\,\,do \Rightarrow  - m + 2 \ne 0 \Leftrightarrow m \ne 2\,\,\\
a)\,\,\,\,Ta\,\,co:\,\,\,d \cap Ox = \left\{ A \right\} \Rightarrow A\left( {\frac{{m - 2}}{m};\,\,\,0} \right)\\
d \cap Oy = \left\{ B \right\} \Rightarrow B\left( {0; - m + 2} \right).\\
 + )\,\,\,\Delta AOB\,\,co\,\,\,S = 2\\
 \Leftrightarrow \frac{1}{2}OA.OB = 2\\
 \Leftrightarrow OA.OB = 4\\
 \Leftrightarrow \left| {{x_A}} \right|.\left| {{y_B}} \right| = 9\\
 \Leftrightarrow \left| {\frac{{ - m + 2}}{m}} \right|.\left| { - m + 2} \right| = 4 \Leftrightarrow \frac{{\left| {m - 2} \right|}}{{\left| m \right|}}.\left| {m - 2} \right| = 4\\
 \Leftrightarrow {\left( {m - 2} \right)^2} = 4\left| m \right|\\
 \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
m > 0\\
{\left( {m - 2} \right)^2} = 4m
\end{array} \right.\\
\left\{ \begin{array}{l}
m < 0\\
{\left( {m - 2} \right)^2} =  - 4m
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
m > 0\\
{m^2} - 8m + 4 = 0
\end{array} \right.\\
\left\{ \begin{array}{l}
m < 0\\
{m^2} + 4 = 0
\end{array} \right.
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
m > 0\\
\left[ \begin{array}{l}
m = 4 + 2\sqrt 3 \\
m = 4 - 2\sqrt 3 
\end{array} \right.
\end{array} \right.\\
\left\{ \begin{array}{l}
m < 0\\
vo\,\,\,nghiem
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
m = 4 + 2\sqrt 3 \,\,\,\,\left( {tm} \right)\\
m = 4 - 2\sqrt 3 \,\,\,\,\left( {tm} \right)
\end{array} \right.\\
Vay\,\,m \in \left\{ {4 - 2\sqrt 3 ;\,\,\,\,4 + 2\sqrt 3 } \right\}\,\,\,thi\,\,\,{S_{OAB}} = 2.\\
b)\,\,\,\,\,\Delta AOB\,\,\,la\,\,\,tam\,\,giac\,\,can \Rightarrow \Delta AOB\,\,\,can\,\,\,tai\,\,O\\
 \Rightarrow OA = OB \Leftrightarrow \left| {{x_A}} \right| = \left| {{y_B}} \right|\\
 \Leftrightarrow \left| {\frac{{m - 2}}{m}} \right| = \left| { - m + 2} \right| \Leftrightarrow \frac{{\left| {m - 2} \right|}}{{\left| m \right|}} = \left| {m - 2} \right|\\
 \Leftrightarrow \left| {m - 2} \right|\left( {\frac{1}{{\left| m \right|}} - 1} \right) = 0\\
 \Leftrightarrow 1 - \left| m \right| = 0\,\,\,\,\,\left( {do\,\,\,m - 2 \ne 0} \right)\\
 \Leftrightarrow \left| m \right| = 1 \Leftrightarrow \left[ \begin{array}{l}
m = 1\,\,\,\left( {tm} \right)\\
m =  - 1\,\,\,\left( {tm} \right)
\end{array} \right.\\
Vay\,\,\,m =  \pm 1\,\,\,\,thi\,\,\,\Delta AOB\,\,\,can.
\end{array}\)