Đáp án:
$\begin{array}{l}
a)m = 3\\
\Rightarrow \left\{ \begin{array}{l}
x + 3y = 3\\
3x + 4y = 6
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
3x + 9y = 9\\
3x + 4y = 6
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
3x + 9y - 3x - 4y = 9 - 6\\
x + 3y = 3
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
5y = 3\\
x = 3 - 3y
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
y = \dfrac{3}{5}\\
x = 3 - 3.\dfrac{3}{5} = \dfrac{6}{5}
\end{array} \right.\\
Vậy\,\left( {x;y} \right) = \left( {\dfrac{6}{5};\dfrac{3}{5}} \right)\,khi:m = 3\\
b)\left\{ \begin{array}{l}
x + my = 3\\
mx + 4y = 6
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
mx + {m^2}y = 3m\\
mx + 4y = 6
\end{array} \right.\\
\Rightarrow {m^2}y - 4y = 3m - 6\\
\Rightarrow \left( {{m^2} - 4} \right).y = 3\left( {m - 2} \right)\\
\Rightarrow \left( {m - 2} \right)\left( {m + 2} \right).y = 3\left( {m - 2} \right)\\
\Rightarrow \left( {m - 2} \right).\left( {m + 2} \right) \ne 0\\
\Rightarrow \left\{ \begin{array}{l}
m \ne 2\\
m \ne - 2
\end{array} \right.\\
\Rightarrow y = \dfrac{{3\left( {m - 2} \right)}}{{\left( {m - 2} \right)\left( {m + 2} \right)}} = \dfrac{3}{{m + 2}}\\
\Rightarrow x = 3 - my = 3 - m.\dfrac{3}{{m + 2}} = \dfrac{6}{{m + 2}}\\
Do:\left\{ \begin{array}{l}
x > 1\\
y > 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\dfrac{6}{{m + 2}} > 1\\
\dfrac{3}{{m + 2}} > 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
\dfrac{{6 - m - 2}}{{m + 2}} > 0\\
m + 2 > 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\dfrac{{4 - m}}{{m + 2}} > 0\\
m + 2 > 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
4 - m > 0\\
m + 2 > 0
\end{array} \right.\\
\Rightarrow - 2 < m < 4\\
Vậy\, - 2 < m < 4;m \ne 2
\end{array}$
