（1）由sinα=

3

5

又 0＜α＜

π

2

∴cosα=

4

5

，tanα=

3

4

∴

sin2α+sin2α

cos2α+cos2α

=

sin2α+2sinα•cosα

2cos2α-sin2α

=

tan2α+2tanα

2-tan2α

=

( 3 4 )2+2× 3 4

2-( 3 4 )2

=

33

23

（2）tan（α-

5

4

π)=

tanα-tan 5 4 π

1+tanα•tan 5 4 π

=

tanα-1

1+tanα

=

3 4 -1

1+ 3 4

=-

1

7

=

tanα-tan 5π 4

1-tanαtan 5π 4

=

tanα-1

1-tanα

=

3 4 -1

1- 3 4

=-1