( 1)X^2-4x+a+3=0 x ∈
(2) When a=0, f (x) = x 2-4x+3 is in x∈, and at this time f(x)∈.
1.m=0 does not meet the problem.
2.m> when g(x) monotonically increases at 0, then g( 1)≤- 1 and g(4)≥3, and the solution is m≥6.
3.m< When g(x) monotonically decreases at 0, then the solutions of g(4)≤- 1 and g( 1)≥3 lead to m≤-3.
Comprehensive m≥6 or m≤-3.
(3)
When 1.t≤0, the range of f(x) is [- 1+a, t 2-4t+a+3].
So the solution of t 2-4t+4 = 7-2t gives t =- 1 and t = 3 (s).
2.0<t<2, the range of f(x) is [- 1+a, a+3], so 4=7-2t t=3/2.
3.2≤t≤4, and the range of f(x) is [t 2-4t+a+3, a+3], so -t 2+4t = 7-2t, and there is no real number solution.
T=- 1, or t=3/2.