14. In the sequence $\left\{a_{n} \mid\right.$, $a_{7}=\frac{16}{3}$, when $n \geqslant 2$, $a_{n}=\frac{3 a_{n-1}-1}{7-a_{n-1}+4}$,
(1) Does there exist a natural number $m$, such that when $n>m$, $a_{n}<2$? If it exists, find the value of $m$; if not, explain the reason.
(2) When $n \geqslant 10$, prove that $\frac{a_{n-1}+a_{n}+1}{2}<a_{n}$.
14. (1) From $a_{7}=\frac{16}{3}$ and the recursive formula, we can get $a_{8}-12, a_{4}=-8$.
$$
a_{n}-2=\frac{3 a_{n-1}+4}{7-a_{n-1}}-2=\frac{3\left(a_{n}-1-2\right)}{7-a_{n}-1} \text {, }
$$

Therefore, if $a_{n-1}2$.
In summary, there exists an $m$ that satisfies the condition, and $m=8$.
（2） $a_{10}=-\frac{4}{3}$, it is easy to prove that when $n \geqslant 10$, $a_{n} \geqslant-\frac{4}{3}$.
$$
\begin{array}{l}
a_{n-1}+a_{n+1}-2 a_{n} \\
=\left(\frac{7 a_{n}-4}{a_{n}+3}-a_{n}\right)+\left(\frac{3 a_{n}+4}{7-a_{n}}-a_{n}\right) \\
=\frac{-\left(a_{n}-2\right)^{2}}{a_{n}+3}+\frac{\left(a_{n}-2\right)^{2}}{7-a_{n}}=\frac{2\left(a_{n}-2\right)^{3}}{\left(7-a_{n}\right)\left(a_{n}+3\right)} .
\end{array}
$$

When $n \geqslant 10$, $a_{n}-20,7-a_{n}>0$, so $\frac{a_{n-1}+a_{n}+1}{2}<a_{n}$.