66 If the parabola $C_{m}: y=x^{2}-m x+m+1$ intersects the line segment $A B$ with endpoints $A(0,4), B(4,0)$ at exactly two points, then the range of values for $m$ is $\qquad$ .
$66\left[3, \frac{17}{3}\right]$. The equation of line segment $AB$ is
$$
x+y=4(0 \leqslant x \leqslant 4)
$$

Substitute $y=-x+4$ into
$$
y=x^{2}-m x+m+1,
$$

we get
$$
x^{2}-(m-1) x+(m-3)=0 .
$$

Let $f(x)=x^{2}-(m-1) x+m-3$. The problem is equivalent to the equation (1) having two real roots $x_{1}, x_{2}$, and $0 \leqslant x_{1}0, \\
0 \leqslant \frac{m-1}{2} \leqslant 4, \\
f(0)=m-3 \geqslant 0, \\
f(4)=16-4(m-1)+m-3 \geqslant 0 .
\end{array}\right.
$$

It is easy to solve that $3 \leqslant m \leqslant \frac{17}{3}$.