解:$(1)A=1-\frac {m}{(m+1)(m-1)}÷\frac {m-1+1}{m-1}$
$=1-\frac {m}{(m+1)(m-1)}·\frac {m-1}{m}$
$=1-\frac {1}{m+1}$
$=\frac {m}{m+1}$
$(3)$由$(1)$知,$A=\frac {m}{m + 1}=1-\frac {1}{m + 1}。$
∵$A$为整数,且$m $也为整数,
∴$m + 1=\pm 1。$∴$m = 0$或$m=-2。$
又∵$m≠\pm 1$且$m≠0,$∴$m=-2。$