14. Let $n$ be a positive integer, and define: $f_{n}(x)=\underbrace{f\{f[\cdots f}_{n \uparrow f}(x) \cdots]\}$, given:
$$
f(x)=\left\{\begin{array}{ll}
2(1-x), & (0 \leqslant x \leqslant 1), \\
x-1, & (1<x \leqslant 2) .
\end{array}\right.
$$
(1) Solve the inequality: $f(x) \leqslant x$;
(2) Let the set $A=\{0,1,2\}$, for any $x \in A$, prove: $f_{3}(x)=x$;
(3) Find the value of $f_{2007}\left(\frac{8}{9}\right)$;
(4) If the set $B=\left\{x \mid f_{12}(x)=x, x \in[0,2]\right\}$, prove: $B$ contains at least 8 elements.
14. (1) (1) When $0 \leqslant x \leqslant 1$, from $2(1-x) \leqslant x$, we get $x \geqslant \frac{2}{3}$, hence $\frac{2}{3} \leqslant x \leqslant 1$.
(2) When $1 < x \leqslant 2$, $x-1 \leqslant x$ always holds, so $1 \leqslant x \leqslant 2$.

Combining (1) and (2), the solution set of the original inequality is $\left\{x \left\lvert\, \frac{2}{3} \leqslant x \leqslant 2\right.\right\}$.
(2) Since $f(0)=2, f(1)=0, f(2)=1$, when $x=0$, $f_{3}(0)=f\{f[f(0)]\}=f\{f[2]\}=f\{1\}=0$; when $x=1$, $f_{A}(1)=f\{f[f(1)]\}=f\{f[0]\}=f\{2\}=1$; when $x=2$, $f_{3}(2)=f\{f[f(2)]\}=f\{f[1]\}=f\{0\}=2$.
Therefore, for any $x \in A$, $f_{3}(x)=x$ always holds.
(3) Since $f\left(\frac{8}{9}\right)=2\left(1-\frac{8}{9}\right)=\frac{2}{9}$.
$f_{2}\left(\frac{8}{9}\right)=f\left(f\left(\frac{8}{9}\right)\right)=f\left(\frac{2}{9}\right)=\frac{14}{9}$,
$f_{3}\left(\frac{8}{9}\right)=f\left(f\left(f\left(\frac{8}{9}\right)\right)\right)=f\left(\frac{14}{9}\right)=\frac{14}{9}-1=\frac{5}{9}$,
$f_{1}\left(\frac{8}{9}\right)=f\left(f\left(f\left(f\left(\frac{8}{9}\right)\right)\right)\right)=f\left(\frac{5}{9}\right)=2\left(1-\frac{5}{9}\right)=\frac{8}{9}$.

In general: $f_{\mathrm{ik}+r}\left(\frac{8}{9}\right)=f_{r}\left(\frac{8}{9}\right)(k, r \in \mathbf{N})$.
So $f_{n}\left(\frac{8}{9}\right)$ has a period of 4.
(4) (1) From the analysis of (1), for the variable $\frac{2}{3}$, we have
$$
f\left(\frac{2}{3}\right)=\frac{2}{3} \text{. }
$$

So $f_{n}\left(\frac{2}{3}\right)=\frac{2}{3}$
Thus, $f_{12}\left(\frac{2}{3}\right)=\frac{2}{3}$, indicating that $\frac{2}{3}$ is an element of $B$:
(2) From the analysis of (2), for the variables $0,1,2$, we always have: $f_{3}(x)=x$, so $f_{12}(x)=f_{1 \times 3}(x)=x$. Hence, $0,1,2$ are also elements of $B$;
(3) From the analysis of (3), for the variables $\frac{8}{9}, \frac{2}{9}, \frac{14}{9}, \frac{5}{9}$, $f_{12}(x)=f_{1 \times 3}(x)=x$.
So $\frac{8}{9}, \frac{2}{9}, \frac{14}{9}, \frac{5}{9}$ are also elements of $B$.
Therefore, the set $B$ contains at least $\frac{2}{3}, 0,1,2, \frac{8}{9}, \frac{2}{9}, \frac{14}{9}, \frac{5}{9}$, these 8 elements.
Note that the conditions or conclusions of this problem are not explicitly given, and it is necessary to propose possible conclusions based on the conditions of the problem. The conclusions may be open-ended, meaning there can be more than one correct answer, and it is not required to find all possible answers. For example, in the 4th question, the conclusion is not unique, and there may be others besides these answers, but it is sufficient to find 8.