解：??$(1)$??將點??$A(1，$????$2)$??代入反比例函數(shù)表達式得，??$2=\frac {k_2}1，$????$k_2=2$??
∴反比例函數(shù)表達式為??$y=\frac 2x$??
將點??$B(m，$????$-1)$??代入反比例函數(shù)表達式得，??$m=-2$??
將點??$A(1，$????$2)、$????$B(-2，$????$-1)$??代入一次函數(shù)得??$\begin{cases}k_1+b=2\\-2k_1+b=-1\end{cases} $??
∴一次函數(shù)表達式為??$y=x+1$??
??$ (2)y_2＜y_1＜y_3$??
??$ (3)$??由圖像可得，??$x＞1$??或??$-2＜x＜0$?
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