1
a+b=(cos(3x/2)-sin(x/2),sin(3x/2)-cos(x/2))
所以：|a+b|^2=(cos(3x/2)-sin(x/2))^2+(sin(3x/2)-cos(x/2))^2
=2-2(cos(3x/2)*sin(x/2)+sin(3x/2)*cos(x/2))=2-2sin(2x)=3
即：sin(2x)=-1/2,因：x∈[π/2,π],所以：2x∈[π,2π]
故：2x=7π/6或2x=11π/6,即：x=7π/12或x=11π/12
2
a dot b=-cos(3x/2)*sin(x/2)-sin(3x/2)*cos(x/2)=-sinx(2x)
所以：f(x)=-sinx(2x)+2-2sin(2x)=2-3sin(2x)
在x∈[π/2,π],即：2x∈[π,2π]时,sin(2x)∈[-1,0]
所以：2-3sin(2x)∈[2,5],因：c>f(x),所以c>5