(1)解:∵数列{an}的前n项和为Sn=2n+c(n∈N*),∴a1=S1=2+c,a2=S2-S1=(4+c)-(2+c)=2,a3=S3-S2=(8+c)-(4+c)=4,∵数列{an}为等比数列,∴22=4(2+c),解得c=-1.(2)c=2时,Sn=2n+2,a1=S1=2+2=4,n≥2时,an=Sn-Sn-1=2n-2n-1=2n-1.∴an= 4,n=1 2n?1,n≥2 .