解析：∵tana+1/tana=10/3∴tana=3,或tana=1/3
又π/4＜a＜π/2,∴tana＞1
只取tana=3,此时sina=3cosa,(sina)^2+(cosa)^2=1
得sina=3√10/10,cosa=√10/10,
∴sin2a=2sinacosa=3/5,cos2a=2(cosa)^2-1=-4/5
则sin(2a+π/4)=sin2acos(π/4)+cos2asin(π/4)=√2/2(sin2a+cos2a)
=√2/2(3/5-4/5)=-√2/10