P={x|2x-a0}
= { x | x > b/3}
P∩Q∩N = { x| b/3 < x < a/2,a,b,x ∈ N } ={1}
=> b = 1 or 2
a = 3 or 4
(a,b) = (3,1) or (3,2) or
(4,1) or (4,2)