6. Given sets:
$$
\begin{array}{l}
A=\left\{x \mid x^{2}+x-6>0\right\}, \\
B=\left\{x \mid x^{2}-2 a x+3 \leqslant 0\right\} .
\end{array}
$$

If $a>0$, and $A \cap B$ contains exactly two integers, then the range of values for $a$ is $\qquad$ .
6. $[2.375,2.8)$.

From the problem, we have
$$
A=\{x \mid x \geq 2\} \text {. }
$$

Since $A \cap B$ contains exactly two integers, we have
$$
A \cap B \neq \varnothing \text {. }
$$

Therefore, $B=\left\{x \mid x^{2}-2 a x+3 \leqslant 0\right\}$ has real roots, i.e., $\Delta=4 a^{2}-12>0$
$\Rightarrow a>\sqrt{3}$ or $a<-\sqrt{3}$. Since $a>0$, then $a>\sqrt{3}$.
Thus, $B=\left\{x \mid a-\sqrt{a^{2}-3} \leqslant x \leqslant a+\sqrt{a^{2}-3}\right\}$.
Since $0 < a - \sqrt{a^2 - 3} < 2$ and $a + \sqrt{a^2 - 3} > 3$, we have:
$$
\begin{array}{l}
a - \sqrt{a^2 - 3} < 2 \Rightarrow a - 2 < \sqrt{a^2 - 3} \Rightarrow (a - 2)^2 < a^2 - 3 \Rightarrow 4a - 7 > 0 \Rightarrow a > \frac{7}{4}, \\
a + \sqrt{a^2 - 3} > 3 \Rightarrow \sqrt{a^2 - 3} > 3 - a \Rightarrow a^2 - 3 > (3 - a)^2 \Rightarrow 6a - 12 > 0 \Rightarrow a > 2.
\end{array}
$$
Since $a - \sqrt{a^2 - 3} \geq 1.5$ and $a + \sqrt{a^2 - 3} \leq 3.5$, we have:
$$
\begin{array}{l}
a - \sqrt{a^2 - 3} \geq 1.5 \Rightarrow a - 1.5 \geq \sqrt{a^2 - 3} \Rightarrow (a - 1.5)^2 \geq a^2 - 3 \Rightarrow 3a - 3.25 \geq 0 \Rightarrow a \geq \frac{13}{12}, \\
a + \sqrt{a^2 - 3} \leq 3.5 \Rightarrow \sqrt{a^2 - 3} \leq 3.5 - a \Rightarrow a^2 - 3 \leq (3.5 - a)^2 \Rightarrow 7a - 15.25 \leq 0 \Rightarrow a \leq \frac{61}{28}.
\end{array}
$$
Combining the above, we get:
$$
\begin{array}{l}
\frac{19}{8} \leqslant a < \frac{14}{5} \\
\Rightarrow 2.375 \leqslant a < 2.8 .
\end{array}
$$