(1)数列{an}为等差数列,a3=3,a7=7,可求得其公差d=1,从而可求得an;同理可求得b1,bn.
(2)由(1)可得cn=n•3n,设数列{cn}的前n项的和为Tn,利用错位相减法即可求得Tn.
【解析】
(1)∵{an}为等差数列,a3=3,a7=7,设其公差为d,则d==1,
∴an=a3+(n-3)d=3+n-3=n;
又数列{bn}为等比数列,q=3,=b1•q5=b1•35,
∴b1=3.
∴bn=3•3n-1=3n.
(2)∵cn=an•bn=n•3n,设数列{cn}的前n项的和为Tn,
则Tn=3+2•32+3•33+…+n•3n①
3Tn=1•32+2•33+…+(n-1)•3n+n•3n+1②
①-②得:-2Tn=3+32+33+…+3n-n•3n+1,
∴Tn=-[-n•3n+1]=.