Solve the following inequality over the set of real numbers:

$$
\log _{x-3}(x-1)<2
$$
Given that if $a>1$, then the function $f(x)=\log _{a} x$ is strictly increasing, while if $0<a<1$, it is strictly decreasing. Using this, solve the following inequalities:

1. $\log _{2}(x-1)<\log _{2}(3-x)$, $(1<x<3)$
2. $\log _{0.5}(x-1)>\log _{0.5}(3-x)$, $(1<x<3)$
3. $\log _{2}(x-1)>\log _{2}(x-3)^{2}$, $(x>3 \text{ or } x<3)$

Solutions:
1. Since $2>1$, the function $\log _{2} x$ is strictly increasing. Therefore, $\log _{2}(x-1)<\log _{2}(3-x)$ implies $x-1<3-x$, which gives $x<2$. Combining with the domain $1<x<3$, the solution is $1<x<2$.
2. Since $0.5<1$, the function $\log _{0.5} x$ is strictly decreasing. Therefore, $\log _{0.5}(x-1)>\log _{0.5}(3-x)$ implies $x-1<3-x$, which gives $x<2$. Combining with the domain $1<x<3$, the solution is $1<x<2$.
3. Since $2>1$, the function $\log _{2} x$ is strictly increasing. Therefore, $\log _{2}(x-1)>\log _{2}(x-3)^{2}$ implies $x-1>(x-3)^{2}$, which simplifies to $x-1>x^2-6x+9$, or $x^2-7x+10<0$. Factoring, we get $(x-2)(x-5)<0$, which gives $2<x<5$. Combining with the domain $x>3$ or $x<3$, the solution is $3<x<5$, or $3<x<4$.