1.最小正周期T=π/2定义域为2x+π/4∈(2kπ-π/2,2kπ+π/2)x∈(kπ-3π/8,kπ+π/8)
2.f(α/2)=tan(α+π/4)
=[tanα+tan(π/4)]/[1-tanα*tan(π/4)]
=(tanα+1)(1-tanα)
=(sinα+cosα)/(cosα-sinα)
=2cos2α=2(cos²α-sin²α)
=2(cosα+sinα)(cosα-sinα)
因α（0,四分之π）,所以cosα+sinα>0
所以2(cosα-sinα)²
=1cosα-sinα=±√2/2√2sin(π/4-α)
=±√2/2sin(π/4-α)
=±1/2π/4-α
=±π/6解得α=5π/12(舍去)或π/12
∴α=π/12