(1)f(x)=√3sin²(x+π/4)-cos²x-(1+√3)/2 =√3/2[1-cos(2x+π/2)]-1/2(1+cos2x)+(1+√3)/2 =-√3/2cos(2x+π/2)-1/2cos2x =√3/2sin2x-1/2cos2x =sin(2x-π/6)∴f(x)的最小值为-1 最小正周期T=2π/2=π (2)∵向量m=(1,5)⊥向量n=(1,f(π/4-A))∴5f(π/4-A)-1=0∴sin[2(π/4-A)-π/6]=1/5∴sin(π/3-2A)=1/5∵0