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(1)
a=(2sinx,1),b=(2sin(π/6-x),√3)
f(x)=a●b+1=4sinxsin(π/6-x)+√3+1
=4sinx(-sinxcosπ/6+cosxsinπ/6)+√3+1
=2 sinxcosx-2√3sin²x+√3+1
=sin2x-√3(1-cos2x)+√3+1
=sin2x+√3cos2x+1
=2(1/2sin2x+√3/2cos2x)+1
=2sin(2x+π/3)+1
由2kπ-π/2≤2x+π/3≤2kπ+π/2,k∈Z
得：kπ-5π/12≤x≤kπ+π/12,k∈Z
∴f(x)单调递增区间为
[kπ-5π/12,kπ+π/12],k∈Z
(2)
根据题意g(x)=f(x+φ)-1=2sin[2(x+φ)+π/3]
g(x)=2sin(2x+2φ+π/3)
若g(x)是奇函数
∴g(-x)=-g(x)
∴sin(-2x+2φ+π/3)=-sin(2x+2φ+π/3)
-sin2xcos(2φ+π/3)+cos2xsin(2φ+π/3)
= -sin2xcos(2φ+π/3)-cos2xsin(2φ+π/3)
∴2cos2xsin(2φ+π/3)=0
∵cos2x为变量
∴sin(2φ+π/3)=0
∴2φ+π/3=kπ,k∈Z
∴φ=kπ/2-π/6
∵φ>0,
∴取k=1得φ最小值π/3
2(1)
∵{an}为等差数列,a1+a3=4,a2+a4=6
∴ 2a1+2d=4,2a1+4d=6
解得：d=1,a1=1
∴an=a1+(n-1)d=1+(n-1)*1=n
（2）
Sn=(1+n)n/2
∴1/Sn=2/[n(n+1)]=2[1/n-1/(n+1)]
∴Tn=1/S1+1/S2+.+1/Sn
=2[1-1/2+1/2-1/3+.+1/n-1/(n+1)]
=2[1-1/(n+1)]
=2n/(n+1)
∴Tn*Sn=2n/(n+1)*(1+n)n/2=n²=a²n
