数学高手进哦~！

解：1.由
tan(π/4-x)=-1/3
得
[tan(π/4)-tanx]/[1+tan(π/4)tanx]=-1/3
即
(1-tanx)/(1+tanx)=-1/3,
解得
tanx=2.
sin²（x+π/4)/(2cos²x+sin2x)=(sinxcosπ/4+cosxsinπ/4)^2/(2cos²x+2sinxcosx)
=1/2(sinx+cosx)^2/[2cosx(sinx+cosx)]=(sinx+cosx)/(4cosx)
=1/4*(tanx+1)=1/4*(2+1)=3/4
2.f(x)=2√3sinxcosx+2cos²x
=√3sin2x+1+cos2x
=2sin(2x+π/6)+1
(1).令x∈[0,π],2x+π/举闷6∈[π/6,13π/6]
当
2x+π/6∈[π/6,π/2],即
x∈[0,π/6]时，f(x)单调递增；
当
2x+π/6∈[π/2,3π/2],即
x∈[π/6，2π/3]时，f(x)单调递减；
当
2x+π/6∈[3π/2,13π/6],即
x∈[2π/3，π]时，f(x)单调递增。
综上所述，f(x)在[0,π]上单调增区和宽间为[0,π/6]与[2π/3，π]，
单调减区间为[π/6，2π/3].
(2).由(1)可知，函数f(x)在[0,π/6]上单调递增，故
f(x)∈[f(0),f(π/6)]
即
f(x)∈[2,3]，
所以函数唤答亮f(x)的值域为[2,3].