（1）f﹙x﹚=向量a·向量b=2cos²x+2√3sinxcosx
=√3sin2x+cos2x+1
=2sin(2x+π/6)+1
最小正周期T=2π／2=π
由,2kπ-π/2≤2x+π/6≤2kπ+π/2,
kπ-π/3≤x≤kπ+π/6,递增区间 [kπ-π/3 ,kπ+π/6],k∈Z.
（2）将函数f(x)=2sin(2x+π/6)+1=2sin2(x+π/12)+1的图像向右平移π/12个单位,向下平移1个单位得到y=2sin2x的图像,向量m=(π/12,-1).
另：设向量m=(h,k),于是 y=2sin[2(x-h)+π/6]+1+k=2sin(2x-2h+π/6)+1+k,得到y=2sin2x,
所以,-2h+π/6=0,1+k=0,∴ h=π/12,k=-1,向量m=(π/12,-1).