8. Let these 5 numbers be $0<x_{1} \leqslant x_{2} \leqslant x_{3} \leqslant x_{4} \leqslant x_{5}$.

First, $x_{4}<\frac{1}{2}$, otherwise $x_{4} \geqslant \frac{1}{2}, x_{5} \geqslant \frac{1}{2}$, then $x_{4}+x_{5} \geqslant 1$, which is a contradiction! Hence $x_{5}=\frac{1}{2}$.
Take $x_{1} 、 x_{2}$, there exists $x_{i}(3 \leqslant i \leqslant 5)$, such that $x_{2}+x_{1}+x_{i}=1$.
Take $x_{4} 、 x_{5}$, there exists $x_{j}(1 \leqslant j \leqslant 3)$, such that $x_{4}+x_{5}+x_{j}=1$.
Then $1=x_{1}+x_{2}+x_{i} \leqslant x_{1}+x_{3}+x_{i} \leqslant x_{1}+x_{4}+x_{i} \leqslant x_{1}+x_{4}+x_{5} \leqslant x_{j}+$ $x_{4}+x_{5}=1$.

Thus $x_{1}+x_{2}+x_{i}=x_{1}+x_{3}+x_{i}=x_{1}+x_{4}+x_{i}$, hence $x_{2}=x_{3}=x_{4}=x$.
So $\left\{\begin{array}{l}x_{1}+\frac{1}{2}+x=1, \\ x_{1}+x+x=1 .\end{array}\right.$ (1) or $\left\{\begin{array}{l}x_{1}+\frac{1}{2}+x=1, \\ 2 x+x=1 .\end{array}\right.$
or $\left\{\begin{array}{l}x_{1}+\frac{1}{2}+x=1, \\ 2 x+\frac{1}{2}=1 .\end{array}\right.$
(1) has no solution, from (2) we get $x=\frac{1}{3}, x_{1}=\frac{1}{6}$, from (3) we get $x=\frac{1}{4}, x_{1}=\frac{1}{4}$.

Therefore, $\left(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right)=\left(\frac{1}{2}, \frac{1}{3}, \frac{1}{3}, \frac{1}{3}, \frac{1}{6}\right)$ or $\left(\frac{1}{2}, \frac{1}{4}, \frac{1}{4}\right.$, $\left.\frac{1}{4}, \frac{1}{4}\right)$.
8 There are 5 positive numbers that satisfy the conditions:
(1) One of the numbers is $\frac{1}{2}$;
(2) From these 5 numbers, any 2 numbers taken, in the remaining 3 numbers there must be one number, which, together with the sum of the two numbers taken, equals 1.
Find these 5 numbers.