$解:∵1−\frac{x−2}{2}＜\frac{1+x}{3},$ $∴x＞2.$ $∴x的最小整数解为3.$ $把x=3代入2x−a=3,得6−a=3,∴a=3\ $
$解:(1)解方程组得:x=\frac{m-1}{2},y=\frac{1-m}{4}$ $∴\begin{cases}{\frac{m+1}{2}＜1}\\{\frac{1-m}{4}＜1}\end{cases}$ $解得-3＜m＜1$ $解不等式组得x≥-5且x≤2n-1$ $∴2n-1＜-5$ $∴n＜-2$ $(2)∵−3＜m＜1,n＜−2,$ $∴原式=m+3+1−m−n−2=2−n$
$解:解不等式①得:2x≥-2,解得x≥-1$ $解不等式②的:5x＜6,解得x≤\frac{6}{5}$ $∴不等式组的解集为−1≤x＜\frac{6}{5}\ $ $∴整数解为−1,0,1$
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