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Đáp án:

$\begin{array}{l}
a)m = 3\\
 \Leftrightarrow \left\{ \begin{array}{l}
x + 3y = 9\\
3x - 3y = 4
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
4x = 13\\
x - y = \dfrac{4}{3}
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x = \dfrac{{13}}{4}\\
y = x - \dfrac{4}{3} = \dfrac{9}{3} = 3
\end{array} \right.\\
Vậy\,\left( {x;y} \right) = \left( {\dfrac{{13}}{4};3} \right)\,khi:m = 3\\
b)\left( {x;y} \right) = \left( { - 1;3} \right)\\
 \Leftrightarrow \left\{ \begin{array}{l}
 - 1 + m.3 = 9\\
 - m - 3.3 = 4
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
3m = 10\\
 - m - 9 = 4
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
m = \dfrac{{10}}{3}\\
m =  - 13
\end{array} \right.\\
Vậy\,m \in \emptyset \\
c)\left\{ \begin{array}{l}
x + my = 9\\
mx - 3y = 4
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
mx + {m^2}y = 9m\\
mx - 3y = 4
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
{m^2}y + 3y = 9m - 4\\
x = 9 - my
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
\left( {{m^2} + 3} \right).y = 9m - 4\\
x = 9 - my
\end{array} \right.\\
Do:{m^2} + 3 \ge 3 > 0
\end{array}$

Vậy hệ luôn có nghiệm duy nhất với mọi m

$\begin{array}{l}
d)\left\{ \begin{array}{l}
y = \dfrac{{9m - 4}}{{{m^2} + 3}}\\
x = 9 - m.\dfrac{{9m - 4}}{{{m^2} + 3}} = \dfrac{{4m + 27}}{{{m^2} + 3}}
\end{array} \right.\\
Khi:x - 3y = \dfrac{{28}}{{{m^2} + 3}} - 3\\
 \Leftrightarrow \dfrac{{4m + 27}}{{{m^2} + 3}} - 3.\dfrac{{9m - 4}}{{{m^2} + 3}} = \dfrac{{28}}{{{m^2} + 3}} - 3\\
 \Leftrightarrow \dfrac{{4m + 27 - 27m + 12}}{{{m^2} + 3}} = \dfrac{{28 - 3{m^2} - 9}}{{{m^2} + 3}}\\
 \Leftrightarrow 39 - 23m =  - 3{m^2} + 19\\
 \Leftrightarrow 3{m^2} - 23m + 20 = 0\\
 \Leftrightarrow 3{m^2} - 3m - 20m + 20 = 0\\
 \Leftrightarrow \left( {m - 1} \right)\left( {3m - 20} \right) = 0\\
 \Leftrightarrow m = 1;m = \dfrac{{20}}{3}\\
Vậy\,m = 1;m = \dfrac{{20}}{3}
\end{array}$