f(x)=2cosxsinx+2cos²x
=sin2x+cos2x+1
=sin（2x+π/4)+1
(1)T=2π/2=π
(2)f(x)max=1+1=2
2x+π/4=π/2+2kπ
所以集合为：{ x| x=π/8+kπ（k∈Z）}
(3)已知sinx的递增区间为：{x|-π/2+2kπ≤x≤π/2+2kπ}
-π/2+2kπ≤2x+π/4≤π/2+2kπ
-3π/8+kπ≤x≤π/8+kπ
递增区间为：【-3π/8+kπ,π/8+kπ】