（1）A(>0)=3,周期2π/w=2(6π-π),w=1/5,當x=π時,取wx+φ=π/5+φ=π/2,得φ=3π/10,
此函數解析式為f(x)=3sin(x/5+3π/10)
（2）問題即是否存在實數m,滿足不等式：sin{√[-(m-1)^2+4]/5+3π/10}＞sin[√(-m^2+4)/5+3π/10].
首先,-(m-1)^2+4>=0,-m^2+4>=0
即|m|<=2.|m-1|=2則-1=當-1=-m^2>-(m-1)^2>=-4,
-3π<3π/10=<√[-(m-1)^2+4]/5+3π/10<√(-m^2+4)/5+3π/10<=4/5+3π/10<2π,
f(x)在[-3π,2π]上遞增,sin{√[-(m-1)^2+4]/5+3π/10}當1/22π>4/5+3π/10=>√[-(m-1)^2+4]/5+3π/10>√(-m^2+4)/5+3π/10>=3π/10>-3π,
f(x)在[-3π,2π]上遞增,sin{√[-(m-1)^2+4]/5+3π/10}>sin[√(-m^2+4)/5+3π/10]
所以,存在實數m,滿足不等式：sin{√[-(m-1)^2+4]/5+3π/10}>sin[√(-m^2+4)/5+3π/10],
m的取值范圍為(1/2,2].