(1)由Sn=an可得Sn+1=an+1,两式相减可求得=,再结合a1=2,即可求得数列{an}的通项公式;
(2)利用错位相减法即可求得数列{bn}的前n项和.
【解析】
(1)∵Sn=an(n∈N*),①
∴Sn+1=an+1(n∈N*),②
②-①得:an+1=an+1-an,
∴an+1=an,
∴=,又a1=2,
an=an-1=×an-2=××…×a1
=na1=2n.
(2)∵bn=an•3n,设数列{bn}的前n项和为Tn,
则Tn=2×3+4×32+6×33+…+2n×3n,③
∴3Tn=2×32+4×33+…+2(n-1)×3n+2n×3n+1,④
③-④得:-2Tn=2×3+2×32+…+2×3n-2n×3n+1
∴-Tn=3+32+…+3n-n×3n+1
=-n×3n+1
=(-n)3n+1-,
∴Tn=(n-)3n+1+.