Đáp án:
$\begin{array}{l}
a)A\left( {x;x} \right) \in {d_1}\\
\Rightarrow x = - \dfrac{1}{2}.x + 3\\
\Rightarrow x + \dfrac{1}{2}x = 3\\
\Rightarrow \dfrac{3}{2}x = 3\\
\Rightarrow x = 2\\
\Rightarrow A\left( {2;2} \right)\\
b){d_3}//{d_2}\\
\Rightarrow {d_3}:y = 2x + b\left( {b \ne - 2} \right)\\
\text{Tại}:{d_1}:Khi:y = - 1\\
\Rightarrow - 1 = - \dfrac{1}{2}.x + 3\\
\Rightarrow x = 8\\
\Rightarrow {d_3} \cap {d_1} = \left( {8; - 1} \right)\\
hay\,\left( {8; - 1} \right) \in {d_3}\\
\Rightarrow - 1 = 2.8 + b\\
\Rightarrow b = - 17\left( {tmdk} \right)\\
\text{Vậy}\,{d_3}:y = 2x - 17
\end{array}$