解答:(1)容易求得f(x)=sin(x+π/3)在g(x)上任意一点(x,y),关于直线对称的点是(π/2-x,y)∴ sin(π/2-x+π/3)=y∴y=cos(x-π/3)即 g(x)=cos(x-π/3)(2)-π/2∴ -5π/6∴ g(x)∈(-√3/2,1]令g(x)=t则3t²-mt+1=0① t=0时,m无解。② t≠0时∴ m=3t+1/t 是对勾函数t∈(0,1], m≥2√3t∈(-√3/2,0) m≤-2√3∴ m的范围是m≥2√3或m≤-2√3