(1)a是第一象限角,且cosα＝1/2,得出a=60°
故f（π-a）=f（120°）=√3sin120°+cos120°=1
（2）f（x）=√3sinx+cosx
=2((√3sinx)/2+(cosx)/2))
=2(sinxcos30°+cosxsin30°)
=2sin(x+π/6)
令y=x+π/6,则y∈[π/6,2π/3]
f（y）=2siny,
当y=π/2时有最大值f（π/2）=2
当y=π/6时有最大值f（π/6）=1
故当x∈[0,π/2],f（x）∈[1,2]