Given a sequence of positive integers $\left\{a_{n}\right\}$ satisfying: $a_{2}, a_{3}$ are prime numbers, and for any positive integers $m, n(m<n)$, we have $a_{m+n}=a_{m}+a_{n}+31, \frac{3 n-1}{3 m-1}<\frac{a_{n}}{a_{m}}<\frac{5 n-2}{5 m-2}$. Find the general term formula of $\left\{a_{n}\right\}$.
Let $b_{n}=a_{n}+31$, then
$$
b_{m+n}=b_{m}+b_{n} \quad\left(m(5 m-2)\left(n b_{1}-31\right) & \Rightarrow b_{1}>\frac{155}{2} .
\end{array}\right.
$$

Combining $a_{1}=b_{1}-31$, we know that $a_{1}$ must be a number in the open interval $\left(\frac{155}{2}-31,93-31\right)$.
Considering $a_{1} \in \mathbf{N}^{*}$, then $a_{1} \in\{47,48, \cdots, 61\}, a_{2}=b_{2}-31=2 b_{1}-31=2 a_{1}+31, a_{3}=$ $b_{3}-31=3 b_{1}-31=3 a_{1}+62$.
Let the set of possible values for $a_{1}$ be $A$. We will first exclude some general cases.
Since $a_{2}, a_{3}$ are prime, we know $3 \nmid a_{2}$, so $a_{1} \not \equiv 1(\bmod 3)$, hence $49,52,55,58,61 \notin A$; $2 \nmid a_{3}$, so $2 \nmid a_{1}$, hence $48,50,54,56,60 \notin A$; $5 \nmid a_{2}$, so $a_{1} \equiv 2(\bmod 5)$, hence $47,57 \notin A$.

Thus, $A \subseteq\{51,53,59\}$. When $a_{1}=51$, $a_{1}=133=7 \times 19$, which is a contradiction; when $a_{1}=53$, $a_{3}=221=$ $13 \times 17$, which is a contradiction; when $a_{1}=59$, $a_{2}=149, a_{3}=239$ are both prime. Therefore, $a_{n}=b_{n}-31=n b_{1}-31=$ $n\left(a_{1}+31\right)-31=90 n-31$.
In conclusion, the general term formula for the sequence $\left\{a_{n}\right\}$ is $a_{n}=90 n-31$.