Express 5/8 as the sum of three positive common fractions in every possible way, where the numerators are 1.
We need to find a triplet of natural numbers $a, b, c$ such that

$$
\frac{5}{8}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}
$$

We can assume that $a$ is the smallest, and $c$ is the largest denominator in the triplet (equality is not excluded), i.e.,

$$
(00)
$$

Let's get an idea of the magnitude of $a$. For this, we replace the second and third terms on the right-hand side of (1) with the not smaller $1/a$ and the certainly smaller 0, respectively, based on (3); thus, we obtain two inequalities with one unknown, where $1/a$ is on the larger and smaller side, respectively:

$$
\begin{gathered}
\frac{5}{8}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \leq \frac{1}{a}+\frac{1}{a}+\frac{1}{a}=\frac{3}{a}, \quad \text { hence } \quad a \leq \frac{24}{5}, \quad \text { hence } \quad a>\frac{8}{5}>1
\end{gathered}
$$

Thus, $a$ has three possibilities: $a=2, a=3$, or $a=4$.

1. Let's first find $b$ and $c$ corresponding to $a=4$. From (1),

$$
\frac{1}{b}+\frac{1}{c}=\frac{5}{8}-\frac{1}{4}=\frac{3}{8}
$$

Repeating the first part of the above reasoning for $b$ based on (3),

$$
\frac{3}{8}=\frac{1}{b}+\frac{1}{c} \leq \frac{1}{b}+\frac{1}{b}=\frac{2}{b}, \quad \text { so } \quad b \leq \frac{16}{3}(<6)
$$

On the other hand, now (2) implies that $b$ is at least 4, so $b=4$ or $b=5$. With these, (4) yields $c=8$ and $c=40/7$, respectively, but the latter is not an integer. - Thus, with $a=4$, we get one decomposition:

$$
\frac{5}{8}=\frac{1}{4}+\frac{1}{4}+\frac{1}{8}
$$

2. Now let $a=3$. Then (1) becomes

$$
\frac{1}{b}+\frac{1}{c}=\frac{5}{8}-\frac{1}{3}=\frac{7}{24}
$$

Similarly to the above,

$$
\frac{7}{24}=\frac{1}{b}+\frac{1}{c} \leq \frac{1}{b}+\frac{1}{b}=\frac{2}{b}, \quad \text { so } \quad b \leq \frac{48}{7}<7
$$

Thus, the possibilities for $b$ are $b=3,4,5$, and 6. The corresponding values of $c$ from (5) are: $c=-24, c=24, c=120/11$, and $c=8$. Only the second and fourth values are natural numbers, so we get two decompositions:

$$
\frac{5}{8}=\frac{1}{3}+\frac{1}{4}+\frac{1}{24}=\frac{1}{3}+\frac{1}{6}+\frac{1}{8}
$$

3. Finally, for the case $a=2$, from (1),

$$
\frac{1}{b}+\frac{1}{c}=\frac{5}{8}-\frac{1}{2}=\frac{1}{8}
$$

Here, we can directly examine all possible pairs of $b$ and $c$ by a suitable transformation. Remove the fractions, collect all terms on the right side of the equation, and try to factorize the expression:

$$
8 b+8 c=b c, \quad 0=b c-8 b-8 c=(b-8)(c-8)-64
$$

Thus, 64 must be decomposed into two integer factors in every possible way. For the factors, (2) implies $b-8 \leq c-8$, and the two factors cannot be negative, because then $b-8 \leq -8, b \leq 0$ would result. Thus, the possible decompositions are:

$$
\begin{aligned}
& b-8=1,2,4,8 \\
& c-8=64,32,16,8
\end{aligned}
$$

From these, the following decompositions result:

(IV-VII)

$$
\frac{5}{8}=\frac{1}{2}+\frac{1}{9}+\frac{1}{72}=\frac{1}{2}+\frac{1}{10}+\frac{1}{40}=\frac{1}{2}+\frac{1}{12}+\frac{1}{24}=\frac{1}{2}+\frac{1}{16}+\frac{1}{16}
$$

Thus, there are seven (I)-(VII) possible decompositions of 5/8 in the required form.

Remark. The decomposition

$$
\frac{5}{8}=\frac{1}{3}+\frac{1}{3}-\frac{1}{24}
$$

obtained for $a=3$ shows that if we drop the positivity requirement, the problem has more solutions. By a similar line of reasoning, we find six more such decompositions:

$$
\begin{aligned}
& \frac{5}{8}=\frac{1}{1}+\frac{1}{8}-\frac{1}{2}=\frac{1}{2}+\frac{1}{7}-\frac{1}{56}=\frac{1}{2}+\frac{1}{6}-\frac{1}{24}=\frac{1}{2}+\frac{1}{4}-\frac{1}{8} \\
& \frac{5}{8}=\frac{1}{1}-\frac{1}{4}-\frac{1}{24}=\frac{1}{1}-\frac{1}{6}-\frac{1}{8}
\end{aligned}
$$