Let f (x) = (4-a) x 2-4x+ 1.

When a=4, f(x)= -4x+ 1, which is a straight line and satisfies f(x).

At point a>4, f (x) = (4-a) x 2-4x+ 1 is a parabola with a downward opening, and there are countless integers satisfying f(x)< 0;

When 0

⊿= 16-4(4-a)=4a>； 0, there is a condition that satisfies f (x).

Let f(x)=0, x1= (4-2 √ a)/(8-2a) = (2-√ a)/(4-a), and x2 = (2+√ a)/(4-a).

X2-x 1 =(2+√a)-(2-√a)/(4-a)=(2√a)/(4-a)

There are exactly three integers in the solution set.

3 & lt(2√a)/(4-a)& lt； four

(2√a)/(4-a)& lt； 4 = = >； 2√a & lt； 16-4a== >√a & lt； 8-2a== >3a^2-32a+64>； 0，8/3 < a & lt； four

3 & lt(2√a)/(4-a)= = & gt； 12-4a<2√a = = & gt； 6-2a & lt； √a = = & gt； a^2-8a+ 12<； 0, the solution is 2

∴ take 8/3 < a <; There are exactly three integers in the solution set of satisfiable inequality.

When a & lt=0, f (x) = (4-a) x 2-4x+ 1 is a parabola with an upward opening, the inequality f (x)

To sum up, there are exactly three integers a ∈ 8/3.