答：
1）
最低点Q（-π/6，-2)，则A=2
f(-π/6)=2sin(-wπ/6+b)=-2，sin(-wπ/6+b)=-1，b-wπ/6=-π/2
f(π/12)=2sin(wπ/12+b)=0，wπ/12+b=0
由上两式解得：w=2，b=-π/6
f(x)=2sin(2x-π/6)
2)
f(a+π/12)=2sin[2(a+π/12)-π/6]=3/8
sin2a=3/16，1+sin2a=19/16
(sina+cosa)^2=19/16
a是第三象限角，sina<0，cosa<0
所以：sina+cosa=-√19/4
3)
y=f(x)+m=2sin(2x-π/6)+m=0
0<=x<=π/2，0<=2x<=π，-π/6<=2x-π/6<=5π/6
-1/2<=sin(2x-π/6)<=1
-1<=2sin(2x-π/6)<=2
所以：-1<=-m<=2
解得：-2<=m<=1
F(x)=2sin(2x-π/6)，接下来两问迎刃而解
内容全部来源于网络收集，如有侵权，请联系网站删除：QQ：24596024