Let f(m)=t, then t∈(0, 1].

"For any m∈R, two different N can be found in the interval, so f(m)=g(n)"

It is equivalent to "Equation G (x) = x 2-2ax+2 = t (t ∈ (0, 1)), that is, x 2-2ax-t+2 = 0 has two different real number solutions on x∈"

Equivalent to any value of t∈(0, 1), the following conditions must be met at the same time:

(1) axis of symmetry x = a ∈ (1, 3);

(2)g(a)=-a^2-t+2<； 0, that is, a 2 > 2-t, 2-t∈[ 1, 2], so a 2 ≥ 2, a ≥ 2;

(3)g( 1)=3-2a-t≥0, that is, 2a≤3-t, 3-t ∈ [2,3], so 2a≤2 and a ≤1;

(4)g(3)= 1 1-6a-t≥0, that is, 6a ≤ 1 1-t,1-t ∈ [/kloc-] .

To sum up, there is no real number A that satisfies the problem.