(1)根据等比数列的通项公式可得a1,由等比数列通项公式可得an,根据,可得bn;
(2)由(1)表示出cn,利用错位相减法可求得Tn.
【解析】
(1)∵数列{an}为等比数列,∴a4=a1q3,∴16=a1•23,∴a1=1,
∴an=2n(n∈N*),
∵数列{bn}的前n项和Sn=n2+n,
∴令n=1,b1=2,
当n≥2时,Sn-1=(n-1)2+(n-1),
∴bn=Sn-Sn-1=n2+n-(n-1)2-(n-1)=n+1,
∴{bn}的通项公式为:bn=n+1(n∈N*);
(2)∵cn=an•bn=(n+1)•2n,
∴Tn=2×2+3×22+4×23+…+n×2n-1+(n+1)×2n
2Tn=2×22+3×23+4×24+…+n×2n+(n+1)×2n+1
∴相减得,-Tn=2×2+(3-2)×22+(4-3)×23+…+[(n-1)-n]×2n-(n+1)×2n+1
∴-Tn=4+22+23+…+22-(n+1)×2n+1
=4+-(n+1)×2n+1
=-n×2n+1
∴Tn=n×2n+1;