Đáp án:
$\begin{array}{l}
1)m = 3\\
\Leftrightarrow 3{x^2} - 5x + 1 = 0\\
\Leftrightarrow {x^2} - \dfrac{5}{3}.x + \dfrac{1}{3} = 0\\
\Leftrightarrow {\left( {x - \dfrac{5}{6}} \right)^2} = \dfrac{{13}}{{36}}\\
\Leftrightarrow x = \dfrac{{5 \pm \sqrt {13} }}{6}\\
2)m.{x^2} - \left( {2m - 1} \right).x + m - 2 = 0\\
\Leftrightarrow \Delta > 0\\
\Leftrightarrow {\left( {2m - 1} \right)^2} - 4m\left( {m - 2} \right) > 0\\
\Leftrightarrow 4{m^2} - 4m + 1 - 4{m^2} + 8m > 0\\
\Leftrightarrow 4m > - 1\\
\Leftrightarrow m > - \dfrac{1}{4}\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = \dfrac{{2m - 1}}{m}\\
{x_1}{x_2} = \dfrac{{m - 2}}{m}
\end{array} \right.\\
x_1^2 + x_2^2 = 2018\\
\Leftrightarrow {\left( {{x_1} + {x_2}} \right)^2} - 2{x_1}{x_2} = 2018\\
\Leftrightarrow \dfrac{{{{\left( {2m - 1} \right)}^2}}}{{{m^2}}} - 2.\dfrac{{m - 2}}{m} = 2018\\
\Leftrightarrow \dfrac{{4{m^2} - 4m + 1 - 2{m^2} + 4m}}{{{m^2}}} = 2018\\
\Leftrightarrow 2016{m^2} = 1\\
\Leftrightarrow {m^2} = \dfrac{1}{{2016}}\\
\Leftrightarrow m = \dfrac{{ \pm \sqrt {14} }}{{168}}\left( {tmdk} \right)\\
3)\left\{ \begin{array}{l}
{x_1} + {x_2} = \dfrac{{2m - 1}}{m} = 2 - \dfrac{1}{m} \Leftrightarrow \dfrac{1}{m} = 2 - {x_1} - {x_2}\\
{x_1}{x_2} = \dfrac{{m - 2}}{m} = 1 - \dfrac{2}{m}
\end{array} \right.\\
\Leftrightarrow {x_1}{x_2} = 1 - 2.\left( {2 - {x_1} - {x_2}} \right)\\
\Leftrightarrow 2\left( {{x_1} + {x_2}} \right) - {x_1}{x_2} = 3
\end{array}$
