5. Given the sets
$$
\begin{array}{l}
A=\left\{x \mid x^{2}+2 x-8>0\right\}, \\
B=\left\{x \mid x^{2}-2 a x+4 \leqslant 0\right\} .
\end{array}
$$

If $a>0$, and $A \cap B$ contains exactly one integer, then the range of values for $a$ is . $\qquad$
5. $\left[\frac{13}{6}, \frac{5}{2}\right)$.

From the problem, we know $A=\{x \mid x<2\}$.
Let $f(x)=x^{2}-2 a x+4$. Then the axis of symmetry of $f(x)$ is $x=a>0$.
From $f(-4)=16+8 a+4>0$, we know
$B \cap\{x \mid x<0\} \neq \emptyset$.
Solving $\left\{\begin{array}{l}a \geqslant 2 \\ f(2)=4-4 a+4 \leqslant 0 \\ f(-4)=16+8 a+4>0\end{array}\right.$
we get $\frac{13}{6} \leqslant a<\frac{5}{2}$.
Thus, the range of $a$ is $\left[\frac{13}{6}, \frac{5}{2}\right)$.