(1)
∵a1+a2+a3=a1(1+q+q²)=28
a2+a3+a4=a1q(1+q+q²)=56
∴q=(a2+a3+a4)/(a1+a2+a3)=2
∴a1(1+2+4)=28
∴a1=4
∴{an}的通项公式为：an=4×2^(n-1)=2^(n+1)
(2)
∵bn=lg[4^n×2^(n+1)]=lg[2^(2n)×2^(n+1)]=lg[2^(3n+1)]=(3n+1)lg2
∴Sn=[4+7+11+……+(3n+1)]lg2
={[(3n+1)+4]n/2}lg2
=[(3n²+5n)/2]lg2