解：?$(2)$?由題意知關于?$x$?的方程?$2x-b+1=0$?的?$“$?相反方程?$”$?為?$(b-1)x-2=0$?
由?$2x-b+1=0$?得?$x=\frac {b-1}{2}$?，由?$(b-1)x-2=0$?得?$x=\frac {2}{b-1}$?
∵關于?$x$?的方程?$2x-b+1=0$?與其?$“$?相反方程?$”$?的解都是整數(shù)
∴?$\frac {b-1}{2}$?與?$\frac {2}{b-1}$?都為整數(shù)?$ $?
又∵?$b$?為整數(shù)，∴?$b-1=±2$?
∴當?$b-1=-2$?時，?$b=-1$?；?$ $?當?$b-1=2$?時，?$b=3$?
綜上所述，整數(shù)?$b$?的值為?$-1$?或?$3$?
?$(3)$?將方程?$2m(4x-1)+2=-7nx$?整理，得?$ (8m+7n)x-2m+2=0$?
∵關于?$x$?的方程?$kx+k=0$?與?$2m(4x-1)+2=-7nx$?互為?$“$?相反方程?$”$?
∴?$k=-(-2m+2)$?，?$-k=8m+7n$?
∴?$-2m+2=8m+7n$?，∴?$10m+7n=2$?
∴?$m -\frac {1}{2}n-[2(3m-1)+3n] =m -\frac {1}{2}n-(6m-2+3n) $?
?$=m -\frac {1}{2}n-6m+2-3n =-5m -\frac {7}{2}n+2 $?
?$=-\frac {1}{2}(10m+7n)+2 =-\frac {1}{2}×2+2 =1$?
