(40 points) (1) Find all mappings $f: \mathbf{Z}_{+} \rightarrow \mathbf{Z}_{+}$, satisfying
$$
f(f(x))=9 x, f(f(f(x)))=27 x ;
$$
(2) Find all mappings $f: \mathbf{Z}_{+} \rightarrow \mathbf{Z}_{+}$, satisfying
$$
f(f(x))=9 x .
$$
(1) For any positive integer $x$,
$9 f(x)=f(f(f(x)))=27 x \Rightarrow f(x)=3 x$.
Thus, the solution is $f(x)=3 x$.
(2) Let the set
$M=\{x \mid x$ is a positive integer not divisible by 9 $\}$.
Take infinite sets $A$ and $B$ such that
$A \cap B=\varnothing, A \cup B=M$.
Take a bijection $u$ from set $A$ to set $B$.
For any positive integer $x$, let $x=9^{m} t, 9 \nmid t$.
If $t \in A$, take $f(x)=9^{m} u(t)$;
If $t \in B$, take $f(x)=9^{m+1} u^{-1}(t)$.
It is easy to verify that such $f(x)$ satisfies $f(f(x))=9 x$.
Next, we show that all $f(x)$ satisfying $f(f(x))=9 x$ can be represented by the above $(A, B, u)$.
First, we prove: $f$ is injective.
For any different positive integers $x, y$, we have
$f(f(x))=9 x \neq 9 y=f(f(y))$
$\Rightarrow f(x) \neq f(y)$.
Consider the sets $C=\{x \mid 9 \nmid f(x), x \in M\}$,
$D=\{x \mid 9 \mid f(x), x \in M\}$,

The sets $C$ and $D$ satisfy $C \cap D=\varnothing, C \cup D=M$.
For any $x \in C$, take $u(x)=f(x)$.
Since $f(u(x))=f(f(x))=9 x$, then $u(x) \in D$, and $u$ is a mapping from set $C$ to set $D$.
For any different elements $x, y$ in set $C$,
$f(u(x))=9 x \neq 9 y=f(u(y))$
$\Rightarrow u(x) \neq u(y) \Rightarrow u$ is injective.
For any element $x$ in set $D$, let $f(x)=9 t, f(x)=f(f(t))$.
By the injectivity of $f$, we know $f(t)=x$.
If $9 \mid t$, let $t=9 s$, then
$t=9 s=f(f(s))$,
$x=f(f(f(s)))=9 f(s)$,

which contradicts $x \in D$.
Thus, $t \in M$, and $f(t)=x$ is not divisible by 9, so $t \in C, u(t)=x$.
Therefore, $u$ is surjective, and $u$ is a bijection.
Hence, $(C, D, u)$ can yield a mapping $g$ that satisfies the conditions, and it is easy to verify that $f(x)=g(x)$ for any positive integer $x$.

In summary, all $f(x)$ satisfying the conditions correspond to a group $(A, B, u)$, and a group $(A, B, u)$ can be used to construct a function $f(x)$ that meets the requirements as described above.