Đáp án:
$\begin{array}{l}
a)\left\{ \begin{array}{l}
x - my = 2 + m\\
x + y = 2
\end{array} \right.\\
\Rightarrow x + y - \left( {x - my} \right) = 2 - \left( {2 + m} \right)\\
\Rightarrow y + my = - m\\
\Rightarrow \left( {m + 1} \right).y = - m\\
\Rightarrow \left\{ \begin{array}{l}
m + 1 \ne 0\\
y = \dfrac{{ - m}}{{m + 1}}\\
x = 2 - y
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m \ne - 1\\
y = - \dfrac{m}{{m + 1}}\\
x = \dfrac{{2m + 2 + m}}{{m + 1}} = \dfrac{{3m + 2}}{{m + 1}}
\end{array} \right.\\
Vậy\,m \ne - 1;\left( {x;y} \right) = \left( {\dfrac{{3m + 2}}{{m + 1}};\dfrac{{ - m}}{{m + 1}}} \right)\\
b)x + 3y > 0\\
\Rightarrow \dfrac{{3m + 2}}{{m + 1}} - 3.\dfrac{m}{{m + 1}} > 0\\
\Rightarrow \dfrac{{3m + 2 - 3m}}{{m + 1}} > 0\\
\Rightarrow \dfrac{2}{{m + 1}} > 0\\
\Rightarrow m + 1 > 0\\
\Rightarrow m > - 1\\
Vậy\,m > - 1
\end{array}$
