已知0＜α＜π/2,sinα=4/5

∵0＜α＜π/2,sinα=4/5
∴cosα>0
∵(sinα)^2+(cosα)^2=1
∴(cosα)^2=1-(sinα)^2=1-(4/5)^2=9/25
从而 cosα=3/5
又 sin2α=2sinα*cosα=2*4/5*3/5=24/25
cos2α=1－2(sinα)^2=1-2*(4/5)^2=1-2*16/25=-7/25
∴（sin²α+sin2α）/（cos²α+cos2α）=((4/5)^2+24/25)/((3/5)^2-7/25)
=(16/25+24/25)/(9/25-7/25)
=40/25/(2/25)
=20.
（2）
∵tanα=sinα/cosα=4/5/(3/5)=4/3
tan(-5π/4)=-tan(5π/4)=-tan(π+π/4)=-tanπ/4=-1
∴tan（α-5π/4)=(tanα-tan(-5π/4))/(1+tanα*tan(-5π/4))
=(4/3-(-1))/(1+4/3*(-1))
=(4/3+1)/(1-4/3)
=7/3/(-1/3)
=-7.

sinα=4/5
cosa = 3/5
tan a = 4/3
（1） （sin²α+sin2α）/（cos²α+cos2α）
= (16/25 + 2sina cosa) / ( 9/25+ 2cos^2a -1)
= 20
(2) tan（α-5π/4)
= （tana - tan 5π/4) / (1+tana * tan5π/4)
= 1/7