(1)∵对任意的n∈N*点(n, Sn n )均在直线y=3x-2上.∴ Sn n =3n?2,∴Sn=3n2-2n;当n≥2时,an=Sn-Sn-1=(3n2-2n)-[3(n-1)2-2(n-1)]=6n-5 ①;当n=1时,a1=S1=3×12-2=1,适合①式,所以an=6n-5;(2)由(1)知,bn= 3 anan+1 = 1 2 ( 1 6n?5 ? 1 6n+1 ),∴Tn= 1 2 [(1- 1 7 )+( 1 7 - 1 13 )+…+( 1 6n?5 ? 1 6n+1 )]= 1 2 (1? 1 6n+1 )∴ 1 2 (1? 1 6n+1 )≤ m 20 对所有n∈N*都成立,只需 1 2 ≤ m 20 ∴m≥10∴满足要求的最小正整数m为10.