Đáp án:

\(\begin{array}{l}
h)\,m = 3\\
i)m = 2 - \dfrac{{\sqrt 3 }}{3}\\
j)m =  - \dfrac{2}{5}\\
k)\,m =  - 5 \pm 2\sqrt 6 
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
h)\,m - 2 = \tan {45^0} \Leftrightarrow m - 2 = 1 \Leftrightarrow m = 3\\
i)\,m - 2 =  - \tan \left( {{{180}^0} - {{150}^0}} \right) \Leftrightarrow m - 2 =  - \dfrac{{\sqrt 3 }}{3}\\
 \Leftrightarrow m = 2 - \dfrac{{\sqrt 3 }}{3}\\
j)\,DK:\,m \ne 2,m \ne  - 3\\
A\left( {0;m + 3} \right);B\left( {\dfrac{{ - m - 3}}{{m - 2}};0} \right)\,la\,giao\,diem\,cua\,\left( d \right)\,voi\,Oy;Ox\\
 \Rightarrow OA = \left| {m + 3} \right|;OB = \left| {\dfrac{{m + 3}}{{m - 2}}} \right|\\
\dfrac{1}{{O{H^2}}} = \dfrac{1}{{O{A^2}}} + \dfrac{1}{{O{B^2}}} = \dfrac{1}{{{{\left( {m + 3} \right)}^2}}} + \dfrac{{{{\left( {m - 2} \right)}^2}}}{{{{\left( {m + 3} \right)}^2}}}\\
 \Rightarrow \dfrac{{{m^2} - 4m + 5}}{{{m^2} + 6m + 9}} = 1 \Rightarrow {m^2} - 4m + 5 = {m^2} + 6m + 9\\
 \Leftrightarrow 10m =  - 4 \Leftrightarrow m =  - \dfrac{2}{5}\left( {tm} \right)\\
k)\,{S_{OAB}} = \dfrac{1}{2}OA.OB = 2 \Leftrightarrow OA.OB = 4\\
 \Leftrightarrow \left| {m + 3} \right|.\left| {\dfrac{{m + 3}}{{m - 2}}} \right| = 4 \Leftrightarrow {\left( {m + 3} \right)^2} = 4\left| {m - 2} \right|\\
TH1:m > 2 \Leftrightarrow {m^2} + 6m + 9 = 4m - 8 \Leftrightarrow {m^2} + 2m + 17 = 0\left( {VN} \right)\\
TH2:m < 2 \Leftrightarrow {m^2} + 6m + 9 = 8 - 4m \Leftrightarrow {m^2} + 10m + 1 = 0\\
 \Leftrightarrow m =  - 5 \pm 2\sqrt 6 
\end{array}\)