Three. (50 points) In a convex $n$-sided rose garden $(n \geqslant 4)$, there is 1 red rose planted at each of the $n$ vertices. There is a straight path between every two red roses, and these paths do not have the situation of "three lines intersecting at one point."

--- They divide the garden into many non-overlapping regions (triangles, quadrilaterals, ...), and each region has one white rose (or black rose).
(1) Find the expression for the total number of roses $f(n)$ in the rose garden.
(2) Can there be exactly 99 roses in the garden? Explain your reasoning.
Three, (1) Solution 1: The straight paths in the rose garden form a convex $n$-sided polygon and its diagonals $\left(k_{n}\right)$. It is easy to know that there are $C_{n}^{2}-n$ diagonals in the graph. Furthermore, since the diagonals do not have "three lines intersecting at one point", the number of intersection points of the diagonals inside the shape is $\mathrm{C}_{n}^{4}$. Now, let's first find the number of regions $g(n)$ inside the $n$-sided polygon, which can be expressed as
$$
f(n)=g(n)+n \text {. }
$$

Now, remove diagonal $l_{1}$, and let $l_{1}$ intersect with other diagonals at $d_{1}$ points, then the graph $K_{n}$ is reduced by $d_{1}+1$ regions. Let the remaining graph be $G_{1}$.

In $G_{1}$, remove diagonal $l_{2}$, and let $l_{2}$ intersect with other diagonals (excluding $l_{1}$) at $d_{2}$ points in $G_{1}$, then the graph $G_{1}$ is reduced by $d_{2}+1$ regions. Let the remaining graph be $G_{2}$.

Proceeding in this manner, each time a diagonal is removed from the remaining graph. When the last diagonal is removed, it no longer intersects with any remaining diagonals, and the graph is reduced by 1 region, leaving 1 region. Thus, $d_{C_{n}^{2}-n}^{2}+1=1$. Summing up,
$$
\begin{aligned}
g(n) & =\sum_{m=1}^{c_{n}^{2}-n}\left(d_{m}+1\right)+1 \\
& =\sum_{m=1}^{c_{n}^{2}-n} d_{m}+\left(C_{n}^{2}-n\right)+1 \\
& =C_{n}^{4}+C_{n}^{2}-n+1=C_{n}^{4}+C_{n-1}^{2} .
\end{aligned}
$$

Therefore, $f(n)=g(n)+n=\dot{C}_{n}^{4}+\mathrm{C}_{n-1}^{2}+n$.
Solution 2: Connect an arc between every two vertices, which adds $n$ new bow-shaped regions. Imagine transplanting a red rose to each bow-shaped region at the vertices, then the number of roses is the number of regions. This problem is the same as a problem in the 1999 No. 3 issue of the High School Mathematical Olympiad training questions, and a new proof is provided here.
Consider the recursive relationship between $f(k)$ and $f(k+1)$ (as shown in Figure 14).
For the case of $k$ points, when point $A_{k+1}$ is added, $A_{k+1}$ connects to the previous $k$ points with $k$ lines (dashed lines in Figure 14). At this point, the graph has $\mathrm{C}_{k+1}^{4}$ diagonal intersection points, and the number of new intersection points added is $C_{k+1}^{4}-C_{k}^{4}$. Since each line $A_{k+1} A_{i}(1 \leqslant i \leqslant k)$ intersects with the sides or diagonals of the original $k$-sided polygon at $d_{i}$ points $\left(d_{i} \geqslant 0\right)$, these intersection points divide $A_{k+1} A_{i}$ into $d_{i}+1$ non-overlapping small segments. Each such small segment divides the region it is in into two, so the line $A_{k+1} A_{i}$ adds $d_{i}+1$ regions. Summing up,
$$
\begin{array}{c}
f(k+1)-f(k)=\sum_{i=1}^{k}\left(d_{i}+1\right) \\
=\sum_{i=1}^{k} d+k=\left(C_{k+1}^{4}-C_{k}^{4}\right)+k=C_{k}^{3}+C_{k}^{1} \\
\text { Therefore, } f(n)=\sum_{k=5}^{n}(f(k)-f(k-1))+f(4) \\
=\sum_{k=5}^{n}\left(C_{k-1}^{3}+C_{k}^{1} 1\right)+8 \quad(f(4)=8) \\
=\left(C_{n}^{3}+\cdots+C_{3}^{3}\right)-C_{3}^{3}+\left(C_{n}^{1} \quad+\cdots+C_{2}^{1}+\right.
\end{array}
$$

Therefore, $f(n)=\sum_{k=5}^{n}(f(k)-f(k-1))+f(4)$
$$
\begin{aligned}
& \left.\mathrm{C}_{1}^{1}\right)-\mathrm{C}_{3}^{1}-\mathrm{C}_{2}^{1}-\mathrm{C}_{1}^{1}+8 \\
= & \mathrm{C}_{n}^{4}+\mathrm{C}_{n}^{2}+1
\end{aligned}
$$

Note: This problem has the background of Euler's theorem, and the formula $F=E-V+2$ can be directly substituted to solve it.
(2) The general formula can be rewritten as
$$
f(n)=\frac{1}{24}\left(n^{4}-6 n^{3}+23 n^{2}-18 n+24\right)
$$

The problem is transformed into solving the equation
$$
\frac{1}{24}\left(n^{4}-6 n^{3}+23 n^{2}-18 n+24\right)=99
$$

which simplifies to
$$
n^{4}-6 n^{3}+23 n^{2}-18 n-2352=0,
$$

or $(n-8)\left(n^{3}+2 n^{2}+39 n+294\right)=0$
Since $n$ is a positive integer, we have
$$
n^{3}+2 n^{2}+39 n+294>0 \text {. }
$$

Solving this, we get $n=8$
Therefore, when $n=8$, there are exactly 99 roses in the rose garden.
(Luo Zengru, Shaanxi Normal University, 710062)