14. Given the sequence $\left\{a_{n}\right\}$, the odd terms form an arithmetic sequence with the first term 1, and the even terms form a geometric sequence with the first term 2. The sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$ is $S_{n}$, and it satisfies $S_{5}=2 a_{4}+a_{5}, a_{9}=a_{3}+a_{4}$.
(1) Find the general term formula of the sequence $\left\{a_{n}\right\}$;
(2) If $a_{m} a_{m+1}=a_{m+2}$, find the value of the positive integer $m$;
(3) Does there exist a positive integer $m$ such that $\frac{S_{2 m}}{S_{2 m-1}}$ is exactly a term in the sequence $\left\{a_{n}\right\}$? If it exists, find all values of $m$ that satisfy the condition; if not, explain the reason.
(1) Let the common difference of the arithmetic sequence be $d$, and the common ratio of the geometric sequence be $q$, then $a_{1}=1, a_{2}=2, a_{3}=1+d, a_{4}=2 q, a_{9}=1+4 d$. Since $S_{5}=2 a_{4}+a_{5}$, we have $a_{1}+a_{2}+a_{3}=a_{4}$, which means $4 d=2 q$. Also, $a_{9}=a_{3}+a_{4}$, so $1+4 d=1+d+2 q$. Solving these equations, we get $d=2, q=3$. Therefore, for $k \in \mathbf{N}^{*}$, we have $a_{2 k-1}=1+(k-1) \cdot 2=2 k-1, a_{2 k}=2 \cdot 3^{k-1}$, hence
$$
a_{n}=\left\{\begin{array}{ll}
n & , n=2 k-1 \\
2 \cdot 3^{\frac{n}{2}-1} & , n=2 k
\end{array} \quad k \in \mathbf{N}^{*}\right.
$$
(2) If $a_{m}=2 k$, then from $a_{m} a_{m+1}=a_{m+2}$, we get $2 \cdot 3^{k-1}(2 k+1)=2 \cdot 3^{k}$, solving which gives $k=1$, so $m=2$; if $a_{m}=2 k-1$, then from $(2 k-1) \cdot 2 \cdot 3^{k-1}=2 k+1$, the left side is even while the right side is odd, which is not valid. Therefore, the positive integer that satisfies the condition is 2.
10 points
(3) For $k \in \mathbf{N}^{*}$, we have
$$
\begin{aligned}
S_{2 k} & =\frac{(1+2 k-1) k}{2}+\frac{2\left(1-3^{k}\right)}{1-3}=k^{2}-1+3^{k} \\
S_{2 k-1} & =S_{2 k}-a_{2 k}=k^{2}-1+3^{k}-2 \cdot 3^{k-1}=k^{2}-1+3^{k-1}
\end{aligned}
$$

Assume there exists a positive integer $m$ such that $\frac{S_{2 m}}{S_{2 m-1}}$ is exactly one term in the sequence $\left\{a_{n}\right\}$. From (1), we know that every term in the sequence is positive, so we can set $\frac{S_{2 m}}{S_{2 m-1}}=L\left(L \in \mathbf{N}^{*}\right)$, then $\frac{m^{2}-1+3^{m}}{m^{2}-1+3^{m-1}}=L$, transforming which gives
$$
(3-L) 3^{m-1}=(L-1)\left(m^{2}-1\right)
$$
$\cdots \cdots \cdots 15$ points

Since $m \geq 1, L \geq 1,3^{m-1}>0$, we have $L \leq 3$, and since $L \in \mathbf{N}^{*}$, $L$ can be $1,2,3$.
When $L=1$, $(3-L) 3^{m-1}>0,(L-1)\left(m^{2}-1\right)=0$, so (1) does not hold;
When $L=2$, $(3-2) 3^{m-1}=(2-1)\left(m^{2}-1\right)$, which means $3^{m-1}=m^{2}-1$.
If $m=1,3^{m-1} \neq m^{2}-1$, let $T_{m}=\frac{m^{2}-1}{3^{m-1}}\left(m \in \mathbf{N}^{*}, m \geq 2\right)$, then
$$
T_{m+1}-T_{m}=\frac{(m+1)^{2}-1}{3^{m}}-\frac{m^{2}-1}{3^{m-1}}=\frac{-2 m^{2}+2 m+3}{3^{m}}=\frac{-2\left(m+\frac{1}{2}\right)^{2}+\frac{7}{2}}{3^{m}} \leq \frac{-2 m^{2}+2 \times 2+3}{3^{2}}T_{3}>\cdots$, so only $T_{2}=1$, at this time $m=2, L=2=a_{2}$.
When $L=3$, $(3-3) 3^{m-1}=(3-1)\left(m^{2}-1\right)$, so $m=1, L=3=a_{3}$.
In summary, there exists a positive integer $m=1$ such that $S_{2} / S_{1}$ is exactly the third term in the sequence $\left\{a_{n}\right\}$; there exists a positive integer $m=2$ such that $S_{4} / S_{3}$ is exactly the second term in the sequence $\left\{a_{n}\right\}$.
20 points