Đáp án:
$\begin{array}{l}
\left\{ \begin{array}{l}
2x + 3y = 3\\
\left( {m + 3} \right).x + 2y = m + 4
\end{array} \right.\\
a)m = - 2\\
\Leftrightarrow \left\{ \begin{array}{l}
2x + 3y = 3\\
x + 2y = 2
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
2x + 3y = 3\\
2x + 4y = 4
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
y = 1\\
x = 2 - 2y = 0
\end{array} \right.\\
Vậy\,\left( {x;y} \right) = \left( {0;1} \right)\,khi:m = - 2\\
b)\left\{ \begin{array}{l}
2x + 3y = 3\\
\left( {m + 3} \right).x + 2y = m + 4
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
4x + 6y = 6\\
3\left( {m + 3} \right).x + 6y = 3\left( {m + 4} \right)
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left( {3m + 9 - 4} \right).x = 3m + 12 - 6\\
y = \dfrac{{3 - 2x}}{3}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = \dfrac{{3m + 6}}{{3m + 5}}\left( {m\# - \dfrac{5}{3}} \right)\\
y = \dfrac{{3 - 2.\dfrac{{3m + 6}}{{3m + 5}}}}{3} = \dfrac{{m + 1}}{{3m + 5}}
\end{array} \right.\\
Khi:\left\{ \begin{array}{l}
x > 0\\
y > 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\dfrac{{3m + 6}}{{3m + 5}} > 0\\
\dfrac{{m + 1}}{{3m + 5}} > 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
m > - \dfrac{5}{3}\\
m < - 2
\end{array} \right.\\
\left[ \begin{array}{l}
m > - 1\\
m < - \dfrac{5}{3}
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
m > - 1\\
m < - 2
\end{array} \right.\\
Vậy\,m < - 2\,hoac\,m > - 1
\end{array}$
