Solution: (1) No, solve the equation x2+x- 12=0, x 1=3, X2 =-4. | X 1 |+| X2 | = 3+4 = 7 = 2 × 3.5。 (2) existence. The reasons are as follows: ∵x2-6x-27=0 and x2+6x-27=0 are even quadratic equations. ∴ suppose c=mb2+n, when b=-6 and c=-27, -27 = 36m+n. 274 = 0 is an even quadratic equation. When b=3, c =-34× 32. ∴ c=-34b2。 For any integer b, c =-34b2, △=b2-4ac, = 4b2.x =? ∴x 1=-32b，x2 = 12b。 ∴| x 1 |+x2 | = 2 | b |, ∫b is an integer, ∴ is an arbitrary integer b, and c =