答：
f(x)=sin(2x-π/3)-cos(2x+π/6)
=cos(π/3)sin2x-sin(π/3)cos2x-cos(π/6)cos2x+sin(π/6)sin2x
=sin2x-√3cos2x
=2*[(1/2)sin2x-(√3/2)cos2x]
=2sin(2x-π/3)
1）
图像向右平移m个单位得到：f(x)=2sin(2x-2m-π/3)
经过原点：f(0)=2sin(-2m-π/3)=0
所以：2m+π/3=kπ
所以：m=kπ/2-π/6>0
所以：k=1时,m的最小值为π/3
2）
f(x)=2sin(2x-π/3)=-8/5
所以：sin(2x-π/3)=-4/5
所以：sin(2x+π/6-π/2)=-sin[π/2-(2x+π/6)]=-cos(2x+π/6)=-4/5
所以：cos(2x+π/6)=4/5
因为：
-π/12<=x<=5π/12
-π/6<=2x<=5π/12
0<=2x+π/6<=π
所以：
0<=2x+π/6<=π/2
结合恒等式(cosa)^2+(sina)^2=1解得：
sin(2x+π/6)=3/5
所以：
f(x+π/4)
=2sin(2x+π/2-π/3)
=2sin(2x+π/6)
=2*(3/5)
=6/5