1.f (a+x) f (a-x) = 4 2a, so let b = 4 2a, then the real number pair (a, b) exists, and f (x) = 4 x is a function of type (a, b).

2. The function g(x) is a (1, 4) type function, and g (1+x) g (1-x) = 4.

When x belongs to [0,2], 1≤g(x)≤3 holds.

Let x=0, then g (1) 2 = 4, because g (1)>0, so g( 1)=2.

X= 1, g(2)g(0)=4 because 1≤g(2)≤3.

Therefore, 4/3≤g(0)≤4, whereas it was originally 1≤g(0)≤3.

therefore

4/3≤g(0)≤3

And when x belongs to [0, 1], g (x) = x2-m (x-1)+1(m >; 0),

Substituting x=0, there is g(0)=m+ 1.

So 1/3≤m≤2.