f'( 1)=2-a f( 1)=0

The tangent equation at (1, f( 1)): y=(2-a)(x- 1)

The intersection of simultaneous solution and x 2+y 2 = 1: x 1.

( 1，0)((3-4a+a^2)/(5-4a+a^2)，2(-2+a)/(5-4a+a^2)

4/(5-4 * A+A 2) = 2 Solution: a= 1 or a=3.

(2) 1+ 1/x^2-a/x>； 0 x^2-ax+ 1>； 0 △=a^2-4<； = 0 | a | & lt= 2∴-2 & lt； = a & lt=2