（1）f(x)=2√3sin^2(π/4+mx)-2cos^2mx-√3(0＜m＜6)
=√3[1-cos(π/2+2mx)]-cos2mx-1-√3
=√3sin2mx-cos2mx-1
=2sin(2mx-π/6)-1
f(0)=2sin(0-π/6)-1
=-2*1/2-1
=-2
关于点(-π/12,1)成中心对称
(0,-2)对称点为：x=-π/12*2-0=-π/6
y=1*2-(-2)=0
即对称点(-π/6,0)
f(-π/6)=0
2sin[2m(-π/6)-π/6]-1=0
sin(-mπ/3-π/6)=1/2
-mπ/3-π/6=π/6+2kπ
-2m-1=1+12k
2m=-12k-2
m=-6k-1
∵0