1.f(π/9)=tan(π/3+π/4)
=(tanπ/3+tanπ/4)/(1-tanπ/3tanπ/4)
=(√3+1)/(1-√3)
=-√3-2
2.
∵f(a/3+π/4)=2
∴tan(a+3π/4+π/4)
= tan(π+a)=tana=2
sina/cosa=2
sina=2cosa代入sin²a+cos²a=1
cos²a=1/5
∴cos(a-Pai/4)=根号2/2*(cosa+sina)=根号2/2*(-根号5/5-2根号5/5)=-3根号10/10