2. Let the equation concerning $x$ be
$$
a x^{2}+(a+2) x+9 a=0
$$

has two distinct real roots $x_{1}, x_{2}$, and $x_{1}<1<x_{2}$. Then, the range of values for $a$ is $\qquad$.
2. $-\frac{2}{11}<a<0, \\ \Delta=(a+2)^{2}-4 \times a \times 9 a>0, \\ f(1)=a+a+2+9 a0, \\ f(1)=a+a+2+9 a>0 .\end{array}\right.$
Solving the inequality set (1), we know that the range of $a$ in (1) and (2) has no common part, therefore, (1) has no solution;
Solving the inequality set (2) yields $\left\{\begin{array}{l}a>-\frac{2}{11} .\end{array}\right.$,
Therefore, the solution set is $-\frac{2}{11}<a<0$.