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