∴f(-x)=e-x+ex=f(x)，

∴f(x) is an even function on r;

(2) Solution: If the inequality about x mf(x)≤e-x+m- 1 holds (0, +∞),

That is m (ex+e-x-1) ≤ e-x-1,

∫x > 0，

∴ex+e-x- 1>0，

That is, m≤

e？ x？ 1

ex+e？ x？ 1

It remains (0, +∞),

Let t=ex, (t > 1), then m≤

1? t

t2？ t+ 1

It remains unchanged (1, +∞).

∵

1? t

t2？ t+ 1

=-

t？ 1

(t？ 1)2+(t？ 1)+ 1

=-

1

t？ 1+

1

t？ 1

+ 1

≥?

1

three

The equal sign holds if and only if t=2, that is, x=ln2.

∴m≤？

1

three

(3) let g(x)=ex+e-x-a(-x3+3x),

Then g'(x)=ex-e-x+3a(x2- 1),

When x > 1 and g' (x) > 0, that is, the function g(x) monotonically increases on [1, +∞).

Therefore, the minimum value of g(x) at this time is g( 1)=e+

1

e

-2a，

F (x0) < a (-x03+3x0) holds because x0∈[ 1, +∞) exists.

Gu e+

1

e

-2a

1

2

(e+

1

e

),

Let h (x) = x-(e-1) lnx-1,

Then h'(x)= 1-

e？ 1

x

By h ′ (x) =1-

e？ 1

x

=0，x=e- 1，

(1) When 0 < x < e- 1, h' (x) < 0, and the function is monotonically decreasing,

(2) when x > e- 1, h' (x) > 0, and the function is monotonically increasing.

The minimum value of ∴h(x) at (0, +∞) is h(e- 1),

Note that h( 1)=h(e)=0,

When x ∴ (1,e- 1)? (0，e- 1)，h (e- 1) ≤ h (x) < h ( 1) = 0，

When x∈(e- 1, e)? (e- 1，+∞)，h (x) < h (e) = 0，

∴ h (x) < 0 holds for any x∈( 1, e).

①a∈(

1

2

(e+

1

e

)，e)？ (1, e), h (a) < 0, that is, A- 1 < (e- 1) LNA, so AE- 1 > EA- 1,

② when a=e, ae- 1=ea- 1,

③ When a∈(e, +∞), e)? (e- 1, +∞), when A > E- 1, H (A) > H (E) = 0, that is, A- 1 > (E- 1) LNA, thus AE-65438+.