Example 4 (1) In a competition with $n(n>1)$ rounds, a total of $m$ medals were awarded. In the first round, 1 medal and $\frac{1}{7}$ of the remaining $m-1$ medals were awarded, in the second round, 2 medals and $\frac{1}{7}$ of the remaining medals were awarded, and so on, until the $n$-th round, where exactly $n$ medals were awarded and no medals were left. How many rounds did the competition include? How many medals were awarded in total?

(2) Divide a circle into $n$ different sectors $(n \geqslant 2)$, denoted as $S_{1}, S_{2}, \cdots, S_{n}$. Each sector can be painted with any of three colors: red, blue, or white, with the requirement that adjacent sectors must have different colors. How many ways are there to paint the sectors?
Analysis: For the 1st question, the number of medals awarded in each round follows a certain pattern, so we can establish a relationship for the number of medals awarded in each round or the number of medals remaining after each round; For the 2nd question, we need to establish a recursive relationship for the total number of coloring methods and find the initial value, thereby obtaining the total number of coloring methods.

Solution (1) Let the number of medals remaining after $k$ rounds of the competition be $a_{k}$. Since the number of medals awarded in the $k$th round is $k + \frac{1}{7}(a_{n-1} - k)$, we have $a_{k} = a_{k-1} - \left[k + \frac{1}{7}(a_{k-1} - k)\right]$, which simplifies to $a_{k} = \frac{6}{7} a_{k-1} - \frac{6}{7} k$ with $a_{0} = m, a_{n} = 0$. Further transformation gives $a_{k} + 6k - 36 = \frac{6}{7}[a_{k-1} + 6(k-1) - 36]$, thus $a_{n} + 6n - 36 = (a_{0} - 36)\left(\frac{6}{7}\right)^{n} = (m - 36)\left(\frac{6}{7}\right)^{n}$, which means $a_{n} = (m - 36)\left(\frac{6}{7}\right)^{n} - (6n - 36)$. Since $a_{n} = 0$, we have $(m - 36) = (n - 6) \frac{7^{n}}{6^{n-1}}$. Given that $n - 6 < 6^{n-1}$ and $7^{n}$ and $6^{n-1}$ are coprime, and $m, n \in \mathbf{N}_{+}$, we conclude $n = 6, m = 36$. Therefore, the competition consists of 6 rounds, and a total of 36 medals are awarded.
(2) Let the total number of coloring methods be $a_{n} (n \geqslant 2)$. When $n = 2$, first color $S_{1}$, which has 3 coloring methods, and then $S_{2}$ has 2 coloring methods, so $a_{2} = 6$. When $n \geqslant 3$, $S_{1}$ has 3 coloring methods, $S_{2}$ has 2 coloring methods, $S_{3}$ has 2 coloring methods, ..., $S_{n-1}$ has 2 coloring methods, and $S_{n}$ still has 2 coloring methods (regardless of whether $S_{n}$ is the same color as $S_{1}$), thus there are $3 \times 2^{n-1}$ coloring methods. However, these $3 \times 2^{n-1}$ coloring methods can be divided into two categories: one where $S_{n}$ is the same color as $S_{1}$, in which case $S_{n}$ and $S_{1}$ are considered as one sector, and there are $a_{n-1}$ coloring methods; the other where $S_{n}$ is a different color from $S_{1}$, in which case there are $a_{n}$ coloring methods, so we have $a_{n} + a_{n-1} = 3 \times 2^{n-1} (n \geqslant 3)$. Let $p_{n} = \frac{a_{n}}{2^{n}}$, then $2 p_{n} + p_{n-1} = 3 (n \geqslant 3)$, thus $p_{n} - 1 = -\frac{1}{2}(p_{n-1} - 1) (n \geqslant 3), p_{2} = \frac{a_{2}}{2^{2}}$, so $p_{n} - 1 = \left(-\frac{1}{2}\right)^{n-2}(p_{2} - 1) = -\left(-\frac{1}{2}\right)^{n-1}$. Therefore, $p_{n} = 1 - \left(-\frac{1}{2}\right)^{n-1}$, and we get $a_{n} = 2^{n} p_{n} = 2^{n} + (-1)^{n} \cdot 2 (n \geqslant 3)$. But when $n = 2$, the formula also holds, so we have $a_{n} = 2^{n} + (-1)^{n} \cdot 2 (n \geqslant 2)$. Hence, there are $a_{n} = 2^{n} + (-1)^{n} \cdot 2 (n \geqslant 2)$ coloring methods.