解答:解:(1)由题意知{an}是首项为1,公比为3的等比数列,
∴an=3n-1,
∴Sn=
1-3n
1-3
=
3n-1
2
.
(2)设数列{bn}的公差为d,b1=a2=3,b3=S3=13,
∴b3-b1=10=2d,解得公差d=5.
∴bn=5n-2,
∴anbn=(5n-2)3n-1.
∵Tn=3•3°+8•31+13•32+…+(5n-2)3n-1①
3Tn=3×3+8×32+…+(5n-7)•3n-1+(5n-2)•3n②
由①-②得:-2Tn=3•3°+5•(31+32+…+3n-1)-(5n-2)•3n,
∴Tn=-
3
2
+
15
4
(1-3n-1)+
5n-2
2
•3n=
9
4
+3n•(
10n-9
4
).