∵数列a[n]的前n项和为S[n],前n项积为T[n],且T[n]=2^[n(1-n)]
∴a[1]=T[1]=2^[1(1-1)]=1
(2)证明：∵T[n]=2^[n(1-n)]
∴T[n-1]=2^[(n-1)(2-n)]
将上面两式相除,得：a[n]=2^[-2(n-1)]
∴a[n]=(1/4)^(n-1)
∵a[n+1]=(1/4)^n
∴a[n+1]/a[n]=1/4
∴a[n]为等比数列
(3)分析：
倘若：(S[n+1]-a)^2=(S[n+2]-a)*(S[n]-a)对n∈N*都成立
那么：S[n+1]^2-2aS[n+1]+a^2=S[n+2]S[n]-aS[n+2]-aS[n]+a^2
即：S[n+1]^2-2aS[n+1]=S[n+2]S[n]-aS[n+2]-aS[n]
∵S[n]=[1-(1/4)^n]/(1-1/4)=4[1-(1/4)^n]/3
∴S[n+1]=4[1-(1/4)^(n+1)]/3
S[n+2]=4[1-(1/4)^(n+2)]/3
∴16[1-2(1/4)^(n+1)+(1/4)^(2n+2)]/9-8a[1-(1/4)^(n+1)]/3
=16[1-(1/4)^n-(1/4)^(n+2)+(1/4)^(2n+2)]/9-4a[1-(1/4)^(n+2)]/3-4a[1-(1/4)^n]/3
即：16(1/4)^n[1+(1/4)^2-2(1/4)]/9=4a(1/4)^n[1+(1/4)^2-2(1/4)]/3
∴a=4/3
答：存在常数a=4/3.
当常数a=4/3时：
∵S[n]=[1-(1/4)^n]/(1-1/4)=4[1-(1/4)^n]/3=4/3-4(1/4)^n/3
∴S[n+1]=4/3-4(1/4)^(n+1)/3
S[n+2]=4/3-4(1/4)^(n+2)/3
∵(S[n+1]-a)^2=(S[n+1]-4/3)^2=[-4(1/4)^(n+1)/3]^2=16[(1/4)^(2n+2)]/9
而：(S[n+2]-a)*(S[n]-a)
=(S[n+2]-4/3)*(S[n]-4/3)
=[-4(1/4)^n/3][-4(1/4)^(n+2)/3]
=16[(1/4)^(2n+2)]/9
∴(S[n+1]-4/3)^2=(S[n+2]-4/3)*(S[n]-a)对n∈N*都成立
即：存在常数a=4/3,使(S[n+1]-a)^2=(S[n+2]-a)*(S[n]-a)对n∈N*都成立