(1) b2*S2=(b1*q)*(a1+a1+d)=2q*(6+d)=32 => q(6+d)=16
b3*S3=(b1*q^2)*(3a1+3d)=2q^2*(9+3d)=120 => q^2*(3+d)=20
联立解得：d=2 ,q=2（因为等差数列｛an｝的各项均为正数则d>0,故d= -6/5 ,q=10/3舍掉）
则：(1) an=3+2(n-1)=2n+1
bn=2*2^(n-1)=2^n
(2) an*bn=(2n+1)*2^n
则Tn=3*2+5*2^2+.+(2n+1)*2^n
(1/2)Tn=3+5*2+.+(2n+1)*2^(n-1)
(1/2)Tn-Tn=3+2*2+2*2^2+.+2*2^(n-1)-(2n+1)*2^n
(-1/2)Tn=3+2*2*[1-2^(n-1)]/(1-2)-(2n+1)*2^n
=3+2^(n+1)-4-(2n+1)*2^n
Tn=2-2*2^(n+1)+(2n+1)*2^(n+1)
=2+(2n-1)*2^(n+1)
(3)Sn为等差数列｛an｝的前n项和,则Sn=n*a1+n(n-1)/2=n*(n+2)
故1/Sn=1/[n*(n+2)]=1/2[1/n-1/(n+2)]
则：(1/S1)+(1/S2)+…(1/Sn)=1/2[(1-1/3)+(1/2-1/4)+(1/3-1/5)+…+(1/n)-1/(n+2)]
=1/2[1+1/2-1/(n+2)]
=3/4-1/(2n+4)