1)f(x)=2sin(x+π/4)^2-√3cos2x-1=-√3cos2x-[1-2sin(x+π/4)^2]
=-√3cos2x- cos(2x+π/2)
=-√3cos2x+sin2x
=2sin(2x-π/3)
最小值为-2,最大值为2
最小正周期为T=π
2)h(x)=f(x+t)=2sin(2x+2t-π/3)关于点（-π/6,0)对称
则2sin(2×(-π/6)+2t-π/3)=0
故2×(-π/6)+2t-π/3=kπ,t属于（0,π）,
所以t= kπ/2+π/3,k∈Z.
t=π/3或5π/6.