14. (16 points) The number of tourists received by a certain tourist area each month approximately follows a periodic pattern, and thus, the number of people working in tourism services in the $n$-th month, $f(n)$, can be approximately described by the function
$$
f(n)=100[A \cos (\omega n+\alpha)+k]
$$

where $n$ is a positive integer representing the month, $A, k$ are positive integers, $\omega>0$, and $\alpha \in\left(\frac{\pi}{2}, \pi\right)$.

Statistical findings show that the number of people working in tourism services in this area each month has the following characteristics:
(i) The number of people working in tourism services in the same month each year is basically the same;
(ii) The number of people working in tourism services in August, the busiest month, and February, the least busy month, differs by about 400 people;
(iii) In February, the number of people working in tourism services in this area is approximately 100, and it increases month by month until August, when it reaches its peak.
(1) Based on the known information, try to determine an expression for $f(n)$ that meets the conditions.
(2) Generally, when the number of people working in tourism services in the area reaches 400 or more, the area enters its peak tourist season. Determine which months of the year are the peak tourist season for this area. Please provide reasons.
15. (20 points) If a real number $x_{0}$ satisfies $f\left(x_{0}\right)=x_{0}$, then $x=x_{0}$ is called a fixed point of the function $f(x)$. Given
$$
f(x)=x^{3}+a x^{2}+b x+3(a, b \text{ are constants })
$$
has two distinct critical points $x_{1}$ and $x_{2}$. Question: Does there exist a real number pair $(a, b)$ such that both $x_{1}$ and $x_{2}$ are fixed points? Prove your conclusion.
14. According to the rules (i), (ii), (iii), we know that the function is a periodic function with a period of 12.
Thus,
$$
T=\frac{2 \pi}{\omega}=12 \Rightarrow \omega=\frac{\pi}{6} \text {. }
$$

Since the number of people is the highest in August, we have
$$
\frac{\pi}{6} \times 8+\alpha=2 k \pi(k \in \mathbf{Z}) \text {. }
$$

Combining $\alpha \in\left(\frac{\pi}{2}, \pi\right)$, we get $\alpha=\frac{2 \pi}{3}$.
By rule (ii),
$$
\begin{array}{l}
f(n)_{\max }=f(8)=100 A+100 k, \\
f(n)_{\min }=f(2)=-100 A+100 k .
\end{array}
$$

From $f(8)-f(2)=200 A=400$, we get $A=2$.
When $n=2$,
$$
\begin{array}{l}
f(2)=200 \cos \left(\frac{\pi}{6} \times 2+\frac{2 \pi}{3}\right)+100 k=100 \\
\Rightarrow k=3 .
\end{array}
$$

In summary, $f(n)=200 \cos \left(\frac{\pi}{6} n+\frac{2 \pi}{3}\right)+300$ meets the conditions.
(2) According to the conditions,
$$
\begin{array}{l}
200 \cos \left(\frac{\pi}{6} n+\frac{2 \pi}{3}\right)+300 \geqslant 400 \\
\Rightarrow \cos \left(\frac{\pi}{6} n+\frac{2 \pi}{3}\right) \geqslant \frac{1}{2} \\
\Rightarrow 12 k-6 \leqslant n \leqslant 12 k-2(k \in \mathbf{Z}) .
\end{array}
$$

Since $n \in[1,12]\left(n \in \mathbf{Z}_{+}\right)$, when $k=1$, $6 \leqslant n \leqslant 10$.

Therefore, $n=6,7,8,9,10$, which means the months of June, July, August, September, and October are the peak tourist season for the region.