1.f(π/3) + f(π/2）=[asin(π/3 - π/3) + b] + [asin(π/2 - π/3) + b] =(asin0 + b) + [asin(π/6) + b]
=b + (1/2)a + b =7
f(π) - f(0) =[asin(π - π/3) + b] - [asin(0 - π/3) + b] =[asin(2π/3) + b] - [asin(-π/3) + b]
=[(√3/2)a + b] - [(-√3/2)a + b] = (√3)a =2√3
由此解得：a=2 ,b=3
∴f(x)=2sin(x - π/3) + 3
2.增区间:2kπ-π/2