3. Given the sets $A=\left\{x \mid x^{2}+2 x-8>0\right\}, B=\left\{x \mid x^{2}-2 a x+4 \leqslant 0\right\}$, if $a>0$, and $A \cap B$ contains exactly one integer, then the range of values for $a$ is $\qquad$ .
$A=(-\infty,-4) \bigcup(2, \infty), B=\left[x_{1}, x_{2}\right]$, Since $x_{1}+x_{2}=2 a>0, x_{1} x_{2}=4$, it is clear that $02$, thus $\left\{\begin{array}{l}9-6 a+4 \leqslant 0, \\ 16-8 a+4>0\end{array} \Rightarrow \frac{13}{6} \leqslant a<\frac{5}{2}\right.$. Therefore, the range of values for $a$ is $\left[\frac{13}{6}, \frac{5}{2}\right)$.