Three. (25 points) There are five numbers $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ that satisfy the following conditions:
(1) One of the numbers is $\frac{1}{2}$;
(2) From these five numbers, any two numbers taken, there must exist one number among the remaining three such that the sum of this number and the two taken numbers is 1.
Find these five numbers.
Let's assume $a_{1} \leqslant a_{2} \leqslant a_{3} \leqslant a_{4} \leqslant a_{5}$. From condition (2), taking $a_{1} 、 a_{2}$, there must exist $a_{i}$ ( $3,4,5$) such that $a_{1}+a_{2}+a_{i}=1$; taking $a_{4} 、 a_{5}$, there must exist $a_{j}(j=1,2,3)$ such that $a_{j}+a_{4}+a_{5}=1$.
$$
\begin{array}{l}
\text { Hence } 1=a_{1}+a_{2}+a_{i} \leqslant a_{1}+a_{2}+a_{5} \\
\leqslant a_{1}+a_{3}+a_{5} \leqslant a_{1}+a_{4}+a_{5} \\
\leqslant a_{j}+a_{4}+a_{5}=1,
\end{array}
$$

Then all the equalities in the above expression hold, and $a_{2}=a_{3}=a_{4}$.
Let $a_{2}=a_{3}=a_{4}=a$.
Taking $a_{3} 、 a_{4}$, there are two cases:
(i) $a_{1}+a_{3}+a_{4}=1$ or $a_{3}+a_{4}+a_{5}=1$
$\Rightarrow a_{4}=a_{5}$ or $a_{3}=a_{1}$
$\Rightarrow$ These five numbers are $a 、 a 、 a 、 a 、 1-2 a$.
Since $\frac{1}{2}$ is among these five numbers, the five numbers are
$$
\frac{1}{4} 、 \frac{1}{4} 、 \frac{1}{4} 、 \frac{1}{4} 、 \frac{1}{2} \text { or } \frac{1}{2} 、 \frac{1}{2} 、 \frac{1}{2} 、 \frac{1}{2} 、 0 \text {. }
$$
(ii) $a_{2}+a_{3}+a_{4}=1 \Rightarrow a=\frac{1}{3}$.

Since $\frac{1}{2} \in\left\{a_{1}, a_{5}\right\} \Rightarrow a_{5}=\frac{1}{2}, a_{1}=\frac{1}{6}$.
Therefore, these five numbers are $\frac{1}{6}, \frac{1}{3}, \frac{1}{3}, \frac{1}{3}, \frac{1}{2}$.