2. Given the sets
$$
A=\left\{x \mid x^{2}-3 x+2 \leqslant 0\right\}, B=\left\{x \left\lvert\, \frac{1}{x-3}<a\right.\right\} \text {. }
$$

If $A \subseteq B$, then the range of real number $a$ is
$\qquad$
2. $\left(-\frac{1}{2},+\infty\right)$.

It is easy to see that $A=\{x \mid 1 \leqslant x \leqslant 2\}$.
When $x \in A$, $\frac{1}{x-3} \in\left[-1,-\frac{1}{2}\right]$.
Therefore, $a>-\frac{1}{2}$.
Thus, the range of values for $a$ is $\left(-\frac{1}{2},+\infty\right)$.