Ta có: $a(b-2)=3=1.3=3.1=(-1).(-3)=(-3).(-1)$
TH1: $\left \{ {{a=1} \atop {b-2=3}} \right.$ =>$\left \{ {{a=1} \atop {b=5}} \right.$
TH2: $\left \{ {{a=3} \atop {b-2=1}} \right.$ =>$\left \{ {{a=3} \atop {b=3}} \right.$
TH1: $\left \{ {{a=-1} \atop {b-2=-3}} \right.$ =>$\left \{ {{a=-1} \atop {b=-1}} \right.$
TH1: $\left \{ {{a=-3} \atop {b-2=-1}} \right.$ =>$\left \{ {{a=-3} \atop {b=1}} \right.$
Vậy $(a,b)∈${$(1;5);(3;3);(-1;-1);(-3;1)$}