Five. (20 points) Try to find all functions $f(x)$ that satisfy the following conditions:
(1) $f(x)$ is a monotonic function defined on $\mathbf{R}$;
(2) For any real numbers $x, y$,
$$
f(x+y)=f(x) f(y) \text{; }
$$
(3) $f(1)=a(0<a \neq 1)$.
Lemma: Let $A$ and $B$ be constants. If for any $n \in \mathbf{N}_{+}$, we have $|A-B| \leqslant \frac{1}{n}$, then $A=B$.

Proof of the Lemma: Suppose $A \neq B$, then according to the problem,
$|A-B| \leqslant 1$.
Let $|A-B|=\frac{1}{m} (m \geqslant 1)$. Take $n=[m]+1$. Then $|A-B|=\frac{1}{m}>\frac{1}{n}$, which contradicts the assumption:
Therefore, $A=B$.
Now, let's solve the original problem.
According to the problem, when $n$ is an integer, it is easy to prove by induction that
$$
f(n x)=(f(x))^{n} \text {. }
$$

From $f(x) f(0)=f(x+0)=f(x)$ and $f(x)$ is not identically zero, we know $f(0)=1$, and
$$
f(x)=f\left(\frac{x}{2}+\frac{x}{2}\right)=\left(f\left(\frac{x}{2}\right)\right)^{2}>0 .
$$

When $a>1$, since $f(1)=a>1=f(0)$, $f(x)$ is an increasing function on $\mathbf{R}$.
For any $x \in \mathbf{R}$, let $k \leqslant x \leqslant k+1 (k \in \mathbf{Z})$.
Then $f(k) \leqslant f(x) \leqslant f(k+1)$.
By $f(k)=(f(1))^{k}=a^{k}$,
$$
f(k+1)=a^{k+1},
$$

we get $\quad a^{k} \leqslant f(x) \leqslant a^{k+1}$.
Taking the logarithm, we have $k \leqslant \log _{a} f(x) \leqslant k+1$.
Thus, $\left|\log _{a} f(x)-x\right| \leqslant 1$.
Substituting $n x (n \in \mathbf{N}_{+})$ for $x$, we get
$\left|\log _{a} f(n x)-n x\right|=n\left|\log _{a} f(x)-x\right| \leqslant 1$,

which implies $\left|\log _{a} f(x)-x\right| \leqslant \frac{1}{n}$.
By the lemma, $\log _{a} f(x)=x$, i.e., $f(x)=a^{x}$.
When $0<a<1$, similarly, we can prove $f(x)=a^{x}$. Therefore, all functions that satisfy the conditions are $f(x)=a^{x}$.