Question 1 Find all integer pairs $(k, m), m>k \geqslant 0, (m, k)=1$, such that the sequence defined as
$$
x_{0}=\frac{k}{m}, x_{n+1}=\left\{\begin{array}{ll}
\frac{2 x_{n}-1}{1-x_{n}}, & x_{n} \neq 1 ; \\
1, & x_{n}=1
\end{array}\right.
$$

must contain the number 1.
Analysis: Define the Fibonacci sequence $\left\{F_{n}\right\}$:
$$
F_{0}=F_{1}=1, F_{n+2}=F_{n+1}+F_{n}(n=0,1, \cdots) \text {. }
$$

By enumerating a small amount of data, it is not difficult to find that the answer is $(k, m)=\left(F_{2 l}, F_{2 l+1}\right)(l \in \mathbf{N})$.
First, from $x_{n+1}=\frac{2 x_{n}-1}{1-x_{n}} \Rightarrow x_{n}=\frac{x_{n+1}+1}{x_{n+1}+2}$.
We will use mathematical induction.
If it contains the number 1, then there exists some $j$ such that
$$
x_{j}=1=\frac{F_{0}}{F_{1}} \text {. }
$$

Assume $x_{p+1}=\frac{F_{2 i}}{F_{2 i+1}}$, then
$$
x_{p}=\frac{\frac{F_{2 i}}{F_{2 i+1}}+1}{\frac{F_{2 i}}{F_{2 i+1}}+2}=\frac{F_{2 i+2}}{F_{2 i+3}} .
$$

Thus, there must be $x_{0}=\frac{F_{2 l}}{F_{2 l+1}}$.
Therefore, $(k, m)=\left(F_{2 l}, F_{2 l+1}\right)(l \in \mathbf{N})$.
Here, we should note the fact that any two consecutive terms of the Fibonacci sequence are coprime (which can be easily proven by contradiction).

On the other hand, it is easy to verify that for all $l \in \mathbf{N}$, $x_{0}=\frac{F_{2 l}}{F_{2 l+1}}$ can make the sequence contain the number 1.