解：(1)f(x)=√3sinxcosx-sin²x+½=½[√3·(2sinxcosx)+(1-2sin²x)]=½[√3sin(2x)+cos(2x)]=(√3/2)sin(2x)+½cos(2x)=sin(2x)cos(π/6)+cos(2x)sin(π/6)=sin(2x+π/6)最小正周期T=2π/2=π(2)2kπ-π/2≤2x+π/6≤2kπ+π/2，(k∈Z)时，函数单调递增此时，kπ-π/3≤x≤kπ+π/6，(k∈Z)函数的单调递增区间为[kπ-π/3，kπ+π/6]，(k∈Z)(3)x∈[0，π/2]，π/6≤2x+π/6≤7π/6sin(7π/6)≤sin(2x+π/6)≤sin(π/2)-½≤sin(2x+π/6)≤12x+π/6=π/2时，即x=π/6时，函数取得最大值，f(x)max=12x+π/6=7π/6时，即x=π/2时，函数取得最小值，f(x)min=-½