1.f'(x)=2x-2-2(x+ 1)/(x+ 1)^2=2x-2-2/(x+ 1)>； 0 means that the solution of (x+ 1)(x2-2)> 0 is x∈[-√2,-1)∨[√2, ∧].

2.f (x) = f (x)-x 2+3x+a has only one zero on [-1/2,2]. That is f (- 1/2) * f (2) < 0.

F(- 1/2)=2ln2- 1/2+a

F(2)=2-2ln3+a

F(- 1/2)* F(2)=(2ln 2- 1/2+a)(2-2ln 3+a)& lt； 0 a∈( 1/2-2ln2，2ln3-2)

3.| x-3 |-| 2x+ 10 |+x+ 15-2 | a+ 13 |≥0

X ≥ 3,2-2 | a+13 |≥ 0, so | a+13 | ≤1a ∈ [-12,-14].

-5≤x & lt； 3:04-X≥| A+ 13 | A+ 13 |≤9 A∈[-22，-4]

- 13≤x & lt； At -5, x+14 ≥| a+13 || a+13 | ≤ 9 Same as above.

To sum up the above a∈[-22, -4]