Đáp án:
$\begin{array}{l}
a)K\left( {x;\dfrac{1}{2}} \right) = d \cap {d_3}\\
\Rightarrow \dfrac{1}{2} = \dfrac{1}{2}.x + \dfrac{3}{2}\\
\Rightarrow x = - 2\\
\Rightarrow K\left( { - 2;\dfrac{1}{2}} \right) \in d\\
\Rightarrow \dfrac{1}{2} = \left( {m - 3} \right).\left( { - 2} \right) + 4m - 1\\
\Rightarrow - 2m + 6 + 4m - 1 = \dfrac{1}{2}\\
\Rightarrow 2m = - \dfrac{9}{2}\\
\Rightarrow m = - \dfrac{9}{4}\\
Vậy\,m = \dfrac{{ - 9}}{4}\\
b)d \bot {d_4}\\
\Rightarrow \left( {m - 3} \right).\dfrac{1}{2}.\left( {3m - 4} \right) = - 1\\
\Rightarrow \left( {m - 3} \right)\left( {3m - 4} \right) = - 2\\
\Rightarrow 3{m^2} - 4m - 9m + 12 + 2 = 0\\
\Rightarrow 3{m^2} - 13m + 14 = 0\\
\Rightarrow \left( {3m - 7} \right)\left( {m - 2} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
m = \dfrac{7}{3}\\
m = 2
\end{array} \right.\\
Vậy\,m = 2;m = \dfrac{7}{3}
\end{array}$
