13. (20 points) The domain of the function $f(x)$ is the set of real numbers $\mathbf{R}$, and it is known that when $x>0$, $f(x)>0$, and for any $m, n \in \mathbf{R}$, we have
$$
f(m+n)=f(m)+f(n).
$$
(1) Discuss the odd-even property and monotonicity of the function $f(x)$;
(2) Let the sets be
$$
\begin{array}{l}
A=\left\{(x, y) \mid f\left(3 x^{2}\right)+f\left(4 y^{2}\right) \leqslant 24\right\}, \\
B=\{(x, y) \mid f(x)-f(a y)+f(3)=0\}, \\
C=\left\{(x, y) \left\lvert\, f(x)=\frac{1}{2} f\left(y^{2}\right)+f(a)\right.\right\},
\end{array}
$$

and $f(1)=2$. If $A \cap B \neq \varnothing$ and $A \cap C \neq \varnothing$, find the range of the real number $a$.
Three, 13. (1) From the given, let $m=n=0$, we get $f(0)=0$;
then let $m=x, n=-x$, we get
$$
f(-x)=-f(x) \text {. }
$$

Therefore, $f(x)$ is an odd function on $\mathbf{R}$.
Let $x_{1}, x_{2} \in \mathbf{R}$, and $x_{2}>x_{1}$. Then $x_{2}-x_{1}>0$.
Since $f(x)$ is an odd function and from the given, we have
$$
\begin{array}{l}
f\left(x_{2}-x_{1}\right)=f\left(x_{2}\right)+f\left(-x_{1}\right) \\
=f\left(x_{2}\right)-f\left(x_{1}\right)>0 .
\end{array}
$$

Thus, $f(x)$ is an increasing function on $\mathbf{R}$.
(2) Note that,
$$
\begin{array}{l}
f(1)=2, f(m+n)=f(m)+f(n) . \\
\text { Hence } f\left(3 x^{2}+4 y^{2}\right) \leqslant 24=12 f(1)=f(12 \text { ） } \\
\Rightarrow A=\left\{(x, y) \mid 3 x^{2}+4 y^{2} \leqslant 12\right\} . \\
\text { From } f(x)-f(a y)+f(3)=0 \\
\Rightarrow f(x-a y+3)=f(0) \\
\Rightarrow B=\{(x, y) \mid x-a y+3=0\} \text {; } \\
\text { From } f(x)=\frac{1}{2} f\left(y^{2}\right)+f(a) \\
\Rightarrow f(2 x-2 a)=f\left(y^{2}\right) \\
\Rightarrow C=\left\{(x, y) \mid y^{2}=2(x-a)\right\} .
\end{array}
$$

Since $A \cap B \neq \varnothing, A \cap C \neq \varnothing$, we get
$$
|a| \geqslant \frac{\sqrt{15}}{3},-\frac{13}{6} \leqslant a \leqslant 2 \text {. }
$$

Therefore, the range of real number $a$ is
$$
\left[-\frac{13}{6},-\frac{\sqrt{15}}{3}\right] \cup\left[\frac{\sqrt{15}}{3}, 2\right] \text {. }
$$