$\begin{array}{l}
A \in \left( P \right) \Leftrightarrow - 3 = a{.2^2} + b.2 - 3 \Leftrightarrow 4a + 2b = 0\\
B \in \left( P \right) \Leftrightarrow 5 = a.{\left( { - 2} \right)^2} + b.\left( { - 2} \right) - 3 \Leftrightarrow 4a - 2b = 8\\
\Rightarrow \left\{ \begin{array}{l}
4a + 2b = 0\\
4a - 2b = 8
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
a = 1\\
b = - 2
\end{array} \right. \Rightarrow \left( P \right):y = {x^2} - 2x - 3
\end{array}$
Câu b: Sai đề.
\[\begin{array}{l}
A \in \left( P \right) \Leftrightarrow 0 = a{.3^2} + b.3 - 3 \Leftrightarrow 9a + 3b = 3\,\,\left( 1 \right)\\
Dinh\,I\left( { - 1;0} \right) \Leftrightarrow \left\{ \begin{array}{l}
- \dfrac{b}{{2a}} = - 1\\
a.{\left( { - 1} \right)^2} + b.\left( { - 1} \right) - 3 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
b = 2a\,\,\left( 2 \right)\\
a - b = 3\,\,\left( 3 \right)
\end{array} \right.\\
Tu\,\left( 2 \right),\left( 3 \right)\,ta\,co:\left\{ \begin{array}{l}
b = 2a\\
a - b = 3
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
a = - 3\\
b = - 6
\end{array} \right.\\
Thay\,vao\,\left( 1 \right)\,ta\,duoc\,9.\left( { - 3} \right) + 3.\left( { - 6} \right) = - 45 \ne 3\,nen\,vo\,ly.\\
Vay\,khong\,co\,a,b.
\end{array}\]
