(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...