25. The function $f(x)$ defined on the set of real numbers $\mathbf{R}$ satisfies the following 3 conditions:
(1) $f(x)>0$ when $x>0$; (2) $f(1)=2$; (3) For any $m, n \in \mathbf{R}$, $f(m+n)=f(m)+f(n)$. Let the sets be
$$
\begin{array}{l}
A=\left\{(x, y) \mid f\left(3 x^{2}\right)+f\left(4 y^{2}\right) \leqslant 24\right\}, \\
B=\{(x, y) \mid f(x)-f(a y)+f(3)=0\}, \\
C=\left\{(x, y) \left\lvert\, f(x)=\frac{1}{2} f\left(y^{2}\right)+f(a)\right.\right\},
\end{array}
$$

If $A \cap B \neq \varnothing$ and $A \cap C \neq \varnothing$, find the range of the real number $a$.
25. From condition (3), let $m=0$, we get $f(0)=0$; then let $n=-x, m=x$, we get $f(-x)=-f(x)$, so $f(x)$ is an odd function on $\mathbf{R}$.
Let $x_{1}, x_{2} \in \mathbf{R}$, and $x_{2}>x_{1}$, then from (1) and (3) we get
$f\left(x_{2}\right)-f\left(x_{1}\right)=f\left(x_{0}\right)+f\left(-x_{1}\right)=f\left(x_{2}-x_{1}\right)>0$ (since $x_{2}-x_{1}>0$). Therefore, $f(x)$ is an increasing function on $\mathbf{R}$.
From (2) $f(1)=2$ and (3), we get $24=12 f(1)=f(12)$, so
$f\left(3 x^{2}\right)+f\left(4 y^{2}\right) \leqslant 24$ which means $f\left(3 x^{2}+4 y^{2}\right) \leqslant f(12)$,
thus $A=\left\{(x, y) \mid 3 x^{2}+4 y^{2} \leqslant 12\right\}$.
Similarly, from $f(x)-f(a y)+f(3)=0$, we get $f(x-a y+3)=f(0)$, thus $B=\{(x, y) \mid x-a y+3=0\}$.
From $f(x)=\frac{1}{2} f\left(y^{2}\right)+f(a)$, we get $f(2 x-2 a)=f\left(y^{2}\right)$. Thus $C=\left\{(x, y) \mid y^{2}=2(x-a)\right\}$.
From $A \cap B \neq \varnothing$, we know the line $x-a y+3=0$ intersects or is tangent to the ellipse $3 x^{2}+4 y^{2}=12$.
By making the transformation $x=2 x^{\prime} \cdot y=\sqrt{3} y^{\prime}$, we can convert the problem to the line: $2 x^{\prime}-\sqrt{3} a y^{\prime}+3=0$ intersecting the circle: $x^{\prime 2}+y^{\prime 2}=1$. This is equivalent to $\frac{3}{\sqrt{4+3 a^{2}}} \leqslant 1, a^{2} \geqslant \frac{5}{3}$, solving gives $a \leqslant-\frac{\sqrt{15}}{3}$ or $a \geqslant \frac{\sqrt{15}}{3}$.
From $A \cap C \neq \varnothing$ is equivalent to the system of equations
$$
\left\{\begin{array}{l}
3 x^{2}+4 y^{2}=12 \\
y^{2}=2(x-a)
\end{array}\right.
$$

having a solution. This is equivalent to the equation
$$
3 x^{2}+8(x-a)-12=0
$$

having a root $\geqslant a$. According to the distribution of roots of a quadratic equation, we get
$$
f(a) \leqslant 0 \text { or }\left\{\begin{array}{l}
f(a)<0 . \\
-\frac{4}{3} \geqslant a, \\
\Delta=64+12(8 a+12) \geqslant 0 .
\end{array}\right.
$$

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