$a)$ $(1)$ là hàm số bậc nhất khi $a\ne0$
$⇔$$\left \{ {{m-1\ne0} \atop {m+1\ne0}} \right.$
$⇔$$\left \{ {{m\ne1} \atop {m\ne-1}} \right.$
$b)(1)$ là hàm số đồng biến khi $a>0$
$⇔$\(\left[ \begin{array}{l}m-1>0\\m+1>0\end{array} \right.\) hay \(\left[ \begin{array}{l}m-1<0\\m+1<0\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}m>1\\m>-1\end{array} \right.\) hay \(\left[ \begin{array}{l}m<1\\m<-1\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}m>1\\m<-1\end{array} \right.\)
$c)$ Thay $A(1;2)$ vào $(1)$
$\dfrac{m-1}{m+1}+m+2=2(m\ne-1)$
$⇔\dfrac{m-1+m(m+1)}{m+1}=0$
$⇔m-1+m^2+m=0$
$⇔m^2+2m-1=0$
$⇔(m+1)^2=2$
$⇔$\(\left[ \begin{array}{l}m+1=\sqrt[]{2}\\m+1=-\sqrt[]{2}\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}m=\sqrt[]{2}-1\\m=-1-\sqrt[]{2}\end{array} \right.\)