（1）F(X)=SIN(X+π/6)+2SIN^2(x/2)
=SIN(X+π/6)+1-COSX
=SIN(X+π/6)+1-SIN(π/2-X)
=2COS[(X+π/6+π/2-X)/2]*SIN[(X+π/6-π/2+X)/2]+1 和差化积公式
=2COS(2π/3)SIN(X-π/6)+1
=-SIN(X-π/6)+1
X属于[0,π],∴X-π/6属于[-π/6,5π/6],
∴SIN(X-π/6)的值域为[SIN(-π/6),SIN(π/2)],即[-1/2,1]
∴F(X)=-SIN(X-π/6)+1的值域为[0,3/2]
(2) B为三角形内角,∴B∈(0,π)
由F(B)=1=-SIN(B-π/6)+1,解得 B=π/6
由正弦定理得 b/c=sinB/sinC,即1/√3=sin(π/6)/sinC=1/2/sinC
解得sinC=√3/2,∴C=π/3或2π/3,∴A=π-B-C=π/2或π/6
当A=π/2时,a/sinA=b/sinB => a=bsinA/sinB=1*sin(π/2)/sin(π/6)=2
当A=π/6时,a/sinA=b/sinB => a=bsinA/sinB=1*sin(π/6)/sin(π/6)=1
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