函数f(x)＝向量a模的平方＋向量a×向量b
=3cos(x/2)²-√3cos(x/2)sin(x/2)
=3/2*cosx+3/2-√3/2*sinx
=√3cos(x+π/3)+3/2
令-π+2kπ≤x+π/3≤2kπ,k∈Z
解得-4π/3+2kπ≤x≤-π/3+2kπ,k∈Z
所以函数的单调递增区间为[-4π/3+2kπ,-π/3+2kπ],k∈Z
(2)
若f(x)=2,则√3cos(x+π/3)+3/2=2
√3cos(x+π/3)=1/2
cos(x+π/3)=√3/6
∵x属于（0,π/2)
∴(x+π/3)属于（π/3,5π/6)
∴sin(x+π/3)=√33/6
所以cosx=cos[(x+π/3)-π/3]
=cos(x+π/3)cosπ/3+sin(x+π/3)sinπ/3
=√3/6*1/2+√33/6*√3/2
=(√3+3√11)/12