?$解：(1)解方程組，得\begin{cases}{ x=m-3 }\\{ y=-2m-4 }\end{cases}$?
?$∴\begin{cases}{ m-3≤0 }\\{ -2m-4\lt 0 }\end{cases}$?
?$得-2\lt m≤3$?
?$(2)原式=3-m-m-2=1-2m$?
?$(3)由不等式，得(2m+1)x\lt 2m+1，∵解集為x\gt 1$?
?$∴2m+1\lt 0，得m\lt -\frac 12$?
?$∴-2\lt m\lt -\frac 12，又m為整數(shù)$?
?$∴m=-1$?
