\[\begin{array}{l}
\left( d \right):\,\,\,y = \frac{{1 - m}}{{m + 2}}x + \left( {1 - m} \right)\left( {m + 2} \right)\\
a)\,\,\,d \bot d':\,\,\,y = \frac{1}{4}x + 1\\
\Rightarrow \left\{ \begin{array}{l}
\frac{{1 - m}}{{m + 2}}.\frac{1}{4} = - 1\\
m + 2 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
1 - m = - 4m - 8\\
m \ne - 2
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
3m = - 9\\
m \ne - 2
\end{array} \right. \Leftrightarrow m = - 3.\\
b)\,\,Hs\,\,\,DB \Leftrightarrow \frac{{1 - m}}{{m + 2}} > 0\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
1 - m > 0\\
m + 2 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
1 - m < 0\\
m + 2 < 0
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
m < 1\\
m > - 2
\end{array} \right.\\
\left\{ \begin{array}{l}
m > 1\\
m < - 2
\end{array} \right.
\end{array} \right. \Leftrightarrow - 2 < m < 1.
\end{array}\]
