Đáp án:

$\begin{array}{l}
a)O \in \left( d \right)\\
 \Leftrightarrow 0 = \left( {2m - 1} \right).0 + 3 - m\\
 \Leftrightarrow 3 - m = 0\\
 \Leftrightarrow m = 3\\
Vậy\,m = 3\\
b)2y - x = 5\\
 \Leftrightarrow 2y = x + 5\\
 \Leftrightarrow y = \dfrac{1}{2}.x + \dfrac{5}{2}\\
\left( d \right)//y = \dfrac{1}{2}x + \dfrac{5}{2}\\
 \Leftrightarrow \left\{ \begin{array}{l}
2m - 1 = \dfrac{1}{2}\\
3 - m \ne \dfrac{5}{2}
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
2m = \dfrac{3}{2}\\
m \ne 3 - \dfrac{5}{2}
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
m = \dfrac{3}{4}\\
m \ne \dfrac{1}{2}
\end{array} \right.\\
 \Leftrightarrow m = \dfrac{3}{4}\\
Vậy\,m = \dfrac{3}{4}\\
c)\left( d \right) \cap Ox = \left( {2;0} \right)\\
 \Leftrightarrow 0 = \left( {2m - 1} \right).2 + 3 - m\\
 \Leftrightarrow 4m - 2 + 3 - m = 0\\
 \Leftrightarrow 3m =  - 1\\
 \Leftrightarrow m =  - \dfrac{1}{3}\\
Vậy\,m =  - \dfrac{1}{3}
\end{array}$