14. Given $y=5 \sin \left(\frac{2 m+1}{3} \pi x+\frac{\pi}{3}\right)$ (where $m \in \mathbf{N}$), for any real number $a$, the value of $y$ is $\frac{5}{4}$ no less than 4 times and no more than 8 times in the interval $[a, a+3]$. Find the value of $m$,
14. This is a function of the form $y=A \sin (\omega x+\varphi)$. The key to solving this problem is the period, such as how many periods are there in the given interval, and the length of the interval, which are all issues we need to address. Clearly, $T=\frac{2}{2 m+1}$, and the interval length is $(a+3)-a=$ 3. Since $\frac{5}{4}$ is not the maximum value of $y$, the value $\frac{5}{4}$ appears twice in each period, and it appears 4 times in 2 periods, and 8 times in 4 periods. Therefore, $\left\{\begin{array}{l}4 T>3, \\ 2 T \leqslant 3,\end{array}\right.$. That is, $\frac{3}{4}<T \leqslant \frac{3}{2}$,
so $\frac{3}{4}<\frac{6}{2 m+1} \leqslant \frac{3}{2}$, which means $\frac{3}{2} \leqslant m<\frac{7}{2}$,
and since $m \in \mathbf{N}$, we have $m=2$ or 3.