Đáp án:
$\begin{array}{l}
1)\left( d \right):y = \left( {m - 1} \right).x + m + 3\\
\left( d \right)//y = - 2x + 1\\
\Leftrightarrow \left\{ \begin{array}{l}
m - 1 = - 2\\
m + 3\# 1
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m = - 1\\
m\# - 2
\end{array} \right.\\
Vậy\,m = - 1\\
2)\left( {1; - 4} \right) \in \left( d \right)\\
\Leftrightarrow - 4 = \left( {m - 1} \right).1 + m + 3\\
\Leftrightarrow m - 1 + m + 3 = - 4\\
\Leftrightarrow 2m = - 6\\
\Leftrightarrow m = - 3\\
Vậy\;m = - 3
\end{array}$
3) Gọi điểm cố định mà đường thẳng luôn đi qua là $M\left( {x;y} \right)$
$\begin{array}{l}
\Leftrightarrow y = \left( {m - 1} \right).x + m + 3\forall m\\
\Leftrightarrow y = \left( {x + 1} \right).m - x + 3\forall m\\
\Leftrightarrow \left( {x + 1} \right).m = y + x - 3\forall m\\
\Leftrightarrow \left\{ \begin{array}{l}
x + 1 = 0\\
y + x - 3 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = - 1\\
y = 3 - x = 4
\end{array} \right.\\
Vậy\,M\left( { - 1;4} \right)
\end{array}$
4)
$\begin{array}{l}
+ Cho:x = 0 \Leftrightarrow y = m + 3\\
\Leftrightarrow \left( d \right) \cap Oy = B\left( {0;m + 3} \right)\\
\Leftrightarrow OB = \left| {m + 3} \right|\\
+ Cho:y = 0 \Leftrightarrow x = \dfrac{{m + 3}}{{1 - m}}\left( {m\# 1} \right)\\
\Leftrightarrow \left( d \right) \cap Ox = A\left( {\dfrac{{m + 3}}{{1 - m}};0} \right)\\
\Leftrightarrow OA = \left| {\dfrac{{m + 3}}{{1 - m}}} \right|\\
{S_{OAB}} = 1\\
\Leftrightarrow \dfrac{1}{2}.OA.OB = 1\\
\Leftrightarrow OA.OB = 2\\
\Leftrightarrow \left| {\dfrac{{m + 3}}{{1 - m}}} \right|.\left| {m + 3} \right| = 2\\
\Leftrightarrow {\left( {m + 3} \right)^2} = 2\left| {m - 1} \right|\\
\Leftrightarrow \left[ \begin{array}{l}
{m^2} + 6m + 9 = 2m - 2\left( {khi:m > 1} \right)\\
{m^2} + 6m + 9 = - 2m + 2\left( {khi:m < 1} \right)
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
{m^2} + 4m + 11 = 0\left( {vn} \right)\\
{m^2} + 8m + 7 = 0\left( {khi:m < 1} \right)
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
m = - 1\left( {tm} \right)\\
m = - 7\left( {tm} \right)
\end{array} \right.\\
Vậy\,m = - 1;m = - 7
\end{array}$
