(2) Let $\mathbf{N}$ denote the set of natural numbers, and $f: \mathbf{N} \rightarrow \mathbf{N}$ satisfies the conditions: $f(1)=1$, and for any natural number $n$,
$$
\left\{\begin{array}{l}
3 f(n) f(2 n+1)=f(2 n)(1+3 f(n)) \\
f(2 n)<6 f(n)
\end{array}\right.
$$

Find all solutions to the equation
$$
f(k)+f(l)=293, k<l
$$
Given that $3 f(n)$ and $1+3 f(n)$ are coprime, from the known equation
$$
\begin{array}{c}
3 f(n) f(2 n+1)=f(2 n)(1+3 f(n)) \\
3 f(n) \mid f(2 n)
\end{array}
$$

Also, $f(2 n)$, where $a_{m-1}>\cdots>a_{1}$ are non-negative integers, we will prove that
$$
f(n)=3^{a_{1}}+3^{a_{2}}+\cdots+3^{a_{m}}
$$

When $n=1$, (3) is the condition $f(1)=1$.
Assume (3) holds for $n<2 s$, then for $n=2 s$, let $s=2^{a_{1}}+2^{a_{2}}+\cdots+2^{a_{m}}$, then
$$
\begin{array}{c}
f(2 s)=3 f(s)=3\left(3^{a_{1}}+3^{a_{2}}+\cdots+3^{a_{m}}\right)= \\
3^{a_{1}+1}+3^{a_{2}+1}+\cdots+3^{a_{m}+1} \\
2 s=2^{a_{1}+1}+2^{a_{2}+2}+\cdots+2^{a_{m}+1}
\end{array}
$$

Thus, (3) also holds for $n=2 s+1$. For $n=2 s+1$, we have
$$
f(2 s+1)=1+3 f(s)=1+3^{a_{1}+1}+\cdots+3^{a_{m}+1}
$$

And
$$
2 s+1=1+2^{a_{1}+1}+\cdots+2^{a_{m}+1}
$$

Thus, (3) also holds for $n=2 s+1$. This proves that (3) holds for all $n$. Since
$$
293=3^{5}+3^{3}+2 \times 3^{2}+3+2 \times 1
$$

We have
$$
\left\{\begin{aligned}
f(l)= & 3^{5}+3^{3}+3^{2}+3+1,3^{5}+3^{3}+3^{2}+1,3^{5}+3^{2}+3+1 \\
& 3^{5}+3^{2}+1 \\
f(k)= & 3^{2}+1,3^{2}+3+1,3^{3}+3^{2}+1,3^{3}+3^{2}+3+1
\end{aligned}\right.
$$

Thus, the pairs $(k, l)$ that satisfy the problem are $(5,47),(7,45),(13,39),(15,37)$.