f(x)=2√3sinxcosx+2cos²x+m=√3sin2x+1+cos2x +m=2sin(2x+π/6) +m+1.(1)f(x)的最小正周期为T=π.由2kπ - π/2 ≤2x+π/6≤2kπ + π/2,得kπ - π/3 ≤x≤kπ + π/6 (k∈Z),∴f(x)的单调递增区间为 [kπ - ...2+m≤f(x)≤3+m是怎么来的？由π/6 ≤2x+π/6≤π/2，得1≤2sin(2x+π/6)≤2，而f(x)=2sin(2x+π/6) +m+1，所以得2+m≤f(x)≤3+m。