Đáp án:
$\begin{array}{l}
1)\left( d \right):y = \left( {m - 1} \right).x + 2m + 2\\
\left( {3;0} \right) \in \left( d \right)\\
\Rightarrow 0 = \left( {m - 1} \right).3 + 2m + 2\\
\Rightarrow 0 = 3m - 3 + 2m + 2\\
\Rightarrow 5m = 1\\
\Rightarrow m = \dfrac{1}{5}\\
2)a)m = 1\\
\Rightarrow {x^2} - 4x + 4 = 0\\
\Rightarrow {\left( {x - 2} \right)^2} = 0\\
\Rightarrow x = 2\\
Vậy\,khi:m = 1 \Rightarrow x = 2\\
b){x^2} - 2\left( {m + 1} \right)x + 4m = 0\\
\Rightarrow \Delta ' > 0\\
\Rightarrow {\left( {m + 1} \right)^2} - 4m > 0\\
\Rightarrow {\left( {m - 1} \right)^2} > 0\\
\Rightarrow m \ne 1\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2\left( {m + 1} \right)\\
{x_1}{x_2} = 4m\\
{x_1} = 2{x_2}
\end{array} \right.\\
\Rightarrow 3{x_2} = 2\left( {m + 1} \right)\\
\Rightarrow {x_2} = \dfrac{2}{3}\left( {m + 1} \right)\\
\Rightarrow {x_1} = \dfrac{4}{3}\left( {m + 1} \right)\\
\Rightarrow \dfrac{2}{3}\left( {m + 1} \right).\dfrac{4}{3}\left( {m + 1} \right) = 4m\\
\Rightarrow \dfrac{2}{9}{\left( {m + 1} \right)^2} = m\\
\Rightarrow 2{m^2} + 4m + 2 = 9m\\
\Rightarrow 2{m^2} - 5m + 2 = 0\\
\Rightarrow \left( {2m - 1} \right)\left( {m - 2} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
m = \dfrac{1}{2}\left( {tm} \right)\\
m = 2\left( {tm} \right)
\end{array} \right.\\
Vậy\,m = \dfrac{1}{2};m = 2
\end{array}$