解:(1)f(x)=2sin²(π/4+x)-√3cos2x=1-cos(π/2+2x)-√3cos2x=1+sin(2x)-√3cos2x=1+sin(2x-π/3)/2π/4≤x≤π/2,π/2≤2x≤π,π/6≤2x-π/3≤2π/3,1/4≤sin(2x-π/3)/2≤1/2,5/4≤f(x)≤3/2。f(x)的最大值和最小值分别为3/2和5/4。(2)若f(x)≥m,则有f(x)-m<2,5/4≤f(x)-3/4,结合f(x)≥m,有-3/4若f(x)
(x)max=3,f(x)min=2.
.1<m<4.