Đáp án:
$\begin{array}{l}
a)Khi:x = 2\\
\Leftrightarrow y = {x^2} = 4\\
\Leftrightarrow \left( {2;4} \right) \in \left( d \right)\\
\Leftrightarrow 4 = 5.2 - m + 2\\
\Leftrightarrow m = 8\\
Khi:{x^2} = 5x - m + 2\\
\Leftrightarrow {x^2} = 5x - 6\\
\Leftrightarrow {x^2} - 5x + 6 = 0\\
\Leftrightarrow \left( {x - 2} \right)\left( {x - 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x = 3 \Leftrightarrow y = {x^2} = 9
\end{array} \right.\\
Vậy\,m = 8;\left( {3;9} \right)\\
b){x^2} = 5x - m + 2\\
\Leftrightarrow {x^2} - 5x + m - 2 = 0\\
\Leftrightarrow \Delta > 0\\
\Leftrightarrow {5^2} - 4.\left( {m - 2} \right) > 0\\
\Leftrightarrow 25 - 4m + 8 > 0\\
\Leftrightarrow 4m < 33\\
\Leftrightarrow m < \dfrac{{33}}{4}\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 5\\
{x_1}.{x_2} = m - 2
\end{array} \right.\\
Do:{x_1};{x_2} > 0\\
\Leftrightarrow {x_1}.{x_2} > 0\\
\Leftrightarrow m - 2 > 0\\
\Leftrightarrow m > 2\\
2\left( {\dfrac{1}{{\sqrt {{x_1}} }} + \dfrac{1}{{\sqrt {{x_2}} }}} \right) = 3\\
\Leftrightarrow \dfrac{{\sqrt {{x_1}} + \sqrt {{x_2}} }}{{\sqrt {{x_1}{x_2}} }} = \dfrac{3}{2}\\
\Leftrightarrow 2.\left( {\sqrt {{x_1}} + \sqrt {{x_2}} } \right) = 3\sqrt {m - 2} \\
\Leftrightarrow 2.\sqrt {{x_1} + {x_2} + 2\sqrt {{x_1}{x_2}} } = 3\sqrt {m - 2} \\
\Leftrightarrow 4.\left( {5 + 2.\sqrt {m - 2} } \right) = 9\left( {m - 2} \right)\\
\Leftrightarrow 20 + 8\sqrt {m - 2} = 9\left( {m - 2} \right)\\
\Leftrightarrow 9\left( {m - 2} \right) - 8\sqrt {m - 2} - 20 = 0\\
\Leftrightarrow \left( {9\sqrt {m - 2} + 10} \right)\left( {\sqrt {m - 2} - 2} \right) = 0\\
\Leftrightarrow \sqrt {m - 2} = 2\\
\Leftrightarrow m - 2 = 4\\
\Leftrightarrow m = 6\left( {tmdk} \right)\\
Vậy\,m = 6
\end{array}$
