1. Solve the inequality:

$$
\log _{9 x} 3 x+\log _{3 x^{2}} 9 x^{2}<\frac{5}{2}
$$
1. Let's make the substitution $t=3x$. The given inequality reduces to $\log _{3 t} t + \log _{\frac{t^{2}}{3}} t^{2} \frac{1}{u+1} + \frac{2}{2-u} - \frac{5}{2} = \frac{4-2u+4u+4-5(u+1)(2-u)}{2(u+1)(2-u)} = \frac{5u^{2}-3u-2}{2(u+1)(2-u)}$. The numerator of this fraction has zeros at points $\frac{3 \pm \sqrt{9+40}}{10} = \frac{3 \pm 7}{10}$, i.e., at points $u=1$ and $u=-\frac{2}{5}$, positive for $u \in \left(-\infty, -\frac{2}{5}\right) \cup (1, \infty)$, and negative otherwise. The denominator has zeros at points $u=-1$ and $u=2$, positive for $u \in (-1, 2)$, and negative otherwise. Since the fraction must be negative, the numerator and denominator must have opposite signs; the numerator is positive and the denominator is negative for $u \in (-\infty, -1) \cup (2, \infty)$, while the numerator is negative and the denominator is positive for $u \in \left(-\frac{2}{5}, 1\right)$. From this set of solutions, we need to exclude the value $u=0$ (which we noted earlier).

From $u=\log _{t} 3$ we get $t=3^{\frac{1}{u}}$, and then $x=\frac{t}{3}=3^{\frac{1}{u}-1}$. Thus, for $u \in (-\infty, -1)$ we get $x \in \left(\frac{1}{9}, \frac{1}{3}\right)$; for $u \in (2, \infty)$ we get $x \in \left(\frac{1}{3}, \frac{1}{\sqrt{3}}\right)$; for $u \in \left(-\frac{2}{5}, 0\right)$ we get $x \in \left(0, \frac{1}{27 \sqrt{3}}\right)$; for $u \in (0, 1)$ we get $x \in (1, \infty)$. Recalling that $x=\frac{1}{3}$ is also a solution (which we verified at the beginning), taking the union of all obtained results, we conclude that the solution to the given inequality is $x \in \left(0, \frac{1}{27 \sqrt{3}}\right) \cup \left(\frac{1}{9}, \frac{1}{\sqrt{3}}\right) \cup (1, \infty)$.