（1）a•b=sinx•cosx+1×(-

1

2

)=sinxcosx-

1

2

，∵a⊥b，∴a•b=0

即sinx•cosx-

1

2

=0，故sinx•cosx=

1

2

．|a+b|=

(sinx+cosx)2+(1- 1 2 )2

=

1+2sinxcosx+ 1 4

=

3

2

．

（2）f（x）=a•（a-b）=a2-a•b=sin2x+12-sinx•cosx+

1

2

=

3

2

+sin2x-sinx•cosx=

3

2

+

1-cos2x

2

-

sin2x

2

=2-

1

2

(sin2x+cos2x)=2-

2

2

sin(2x+

π

4

)．∵-1≤sin(2x+

π

4

)≤1，

∴2-

2

2

≤2-

2

2

sin(2x+

π

4

)≤2+

2

2

．故函数f（x）=a•（a-b）的值域为[2-

2

2

，2+

2

2

]．