1,
f（x）=a* sin（2wx+π/6）+a/2+b
则最小正周期为 T=2π/2w=π/w=π得w=1,
即f（x）=a* sin（2x+π/6）+a/2+b ,
f（x）的最大值是7/4 ,最小值3/4相差2a=1,
得a=1/2,
f（x）=1/2* sin（2x+π/6）+1/4+b
f(x)max=1/2+1/4+b=7/4得b=1；
2,
f（x）=1/2* sin（2x+π/6）+5/4
f（x）的单调递增区间 即为 sin（2x+π/6）的单调递增区间,
-π/2+2kπ≤2x+π/6≤π/2+2kπ
-2π/3+2kπ≤2x≤π/3+2kπ
-π/3+kπ≤x≤π/6+kπ
即单调递增区间为[-π/3+kπ,π/6+kπ]