Đáp án:
$\begin{array}{l}
a)A\left( {1;2} \right) \in d\\
\Rightarrow 2 = 3m.1 - 2m - \dfrac{1}{4}\\
\Rightarrow m = 2 + \dfrac{1}{4} = \dfrac{9}{4}\\
b){x^2} = 3mx - 2m - \dfrac{1}{4}\\
\Rightarrow {x^2} - 3mx + 2m + \dfrac{1}{4} = 0\\
\Rightarrow \Delta > 0\\
\Rightarrow {\left( {3m} \right)^2} - 4.\left( {2m + \dfrac{1}{4}} \right) > 0\\
\Rightarrow 9{m^2} - 8m - 1 > 0\\
\Rightarrow \left( {m - 1} \right)\left( {9m + 1} \right) > 0\\
\Rightarrow \left[ \begin{array}{l}
m > 1\\
m < - \dfrac{1}{9}
\end{array} \right.\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 3m\\
{x_1}{x_2} = 2m + \dfrac{1}{4}
\end{array} \right.\\
x_1^2 + x_2^2 = 5\\
\Rightarrow {\left( {{x_1} + {x_2}} \right)^2} - 2{x_1}{x_2} = 5\\
\Rightarrow 9{m^2} - 2.\left( {2m + \dfrac{1}{4}} \right) = 5\\
\Rightarrow 9{m^2} - 4m - \dfrac{1}{2} - 5 = 0\\
\Rightarrow \left[ \begin{array}{l}
m = \dfrac{{4 + \sqrt {214} }}{{18}}\left( {tm} \right)\\
m = \dfrac{{4 - \sqrt {214} }}{{18}}\left( {ktm} \right)
\end{array} \right.\\
c)Dk:m > 1/m < - \dfrac{1}{9}\\
{y_1} = x_1^2;{y_2} = x_2^2\\
\Rightarrow {y_1} + {y_2} - 5{y_1}{y_2} = 10\\
\Rightarrow x_1^2 + x_2^2 - 5.x_1^2.x_2^2 = 10\\
\Rightarrow 9{m^2} - 4m - \dfrac{1}{2} - 5.{\left( {2m + \dfrac{1}{4}} \right)^2} = 10\\
\Rightarrow 9{m^2} - 4m - \dfrac{1}{2} - 5.\left( {4{m^2} + m + \dfrac{1}{{16}}} \right) - 10 = 0\\
\Rightarrow 11{m^2} + 9m + \dfrac{{173}}{{16}} = 0\left( {vô\,nghiệm} \right)
\end{array}$
