13. Given: $a_{1}, a_{2}, \cdots, a_{n}$ are all non-negative real numbers, and satisfy $a_{1}+a_{2}+\cdots+a_{n}=1$, when $m \leqslant \sum_{1}^{n} \sqrt{1+3 a}, \quad \leqslant$ always holds, try to find: the maximum value of $m$ and the minimum value of $M$.
13. First, prove a lemma: If $a, b \in \mathbf{R}^{+}$, then $\sqrt{a+1}+\sqrt{b+1}>1+\sqrt{1+a+b}$
In fact, because $(a+1)(b+1)=a \cdot b+1+a+b>1+a+b$,
so $\sqrt{(a+1)(b+1)}>\sqrt{1+a+b}$, thus $a+1+b+1+2\sqrt{(a+1)(b+1)}>a+1+b+1+2\sqrt{1+a+b}$
i.e.,
$$
(\sqrt{a+1}+\sqrt{b+1})^{2}>(1+\sqrt{1+a+b})^{2}
$$

Therefore,
$$
\sqrt{a+1}+\sqrt{b+1}>1+\sqrt{1+a+b}
$$

By repeatedly using the above inequality, we have:
$$
\sum_{i=1}^{n} \sqrt{1+3 a_{i}}>n-1+\sqrt{1+3 \sum_{i=1}^{n} a_{i}}=n+1
$$

When $a_{1} \rightarrow 1, a_{i}(i=2, \cdots, n) \rightarrow 0$, $\sum_{i=1}^{n} \sqrt{1+3 a_{i}} \rightarrow n+1$
Thus, $m \leqslant n+1$.

On the other hand: Let $\sqrt{x}=\sqrt{\frac{n+3}{n}}$, then $\sqrt{1+3 a_{i}} \cdot \sqrt{x} \leqslant \frac{1+3 a_{i}+x}{2}(i=1,2, \cdots, n)$, so $\sum_{i=1}^{n} \sqrt{1+3 a_{i}} \sqrt{x} \leqslant \frac{1}{2} \sum_{i=1}^{n}\left(1+3 a_{i}+x\right)=\frac{1}{2}(n+3+n+3)=n+3$.
When and only when $a_{1}=a_{2}=\cdots=a_{n}$, $\sum_{i=1}^{n} \sqrt{1+3 a_{i}}=\sqrt{n(n+3)}$.
Therefore, $M \geqslant \sqrt{n(n+3)}$.