We know about an 80-term sequence that any intermediate term is equal to the product of its neighbors, and the product of the first 40 terms is 8, as well as the product of all terms is 8. Determine the first and second elements of the sequence!
The product of the elements of the sequence is 8, so 0 is not an element of the sequence. If we denote the $n$-th element of the sequence by $a_{n}$, then according to the condition,

$$
a_{n+2}=\frac{a_{n+1}}{a_{n}}
$$

Using this, the elements of the sequence can be expressed in terms of the first and second elements as follows:

$$
\begin{gathered}
a_{3}=\frac{a_{2}}{a_{1}}, \quad a_{4}=\frac{a_{3}}{a_{2}}=\frac{1}{a_{1}}, \quad a_{5}=\frac{a_{4}}{a_{3}}=\frac{1}{a_{2}} \\
a_{6}=\frac{a_{5}}{a_{4}}=\frac{a_{1}}{a_{2}}, \quad a_{7}=\frac{a_{6}}{a_{5}}=a_{1}, \quad a_{8}=\frac{a_{7}}{a_{6}}=a_{2}
\end{gathered}
$$

Since $a_{7}=a_{1}$ and $a_{8}=a_{2}$, the first six elements of the sequence will repeat in the same order from then on, i.e., $a_{n+6}=a_{n}$. Furthermore, since the product of the first six elements is 1, the product of the first $6k$ elements is also 1 for any $k$. Therefore, the product of the first 40 elements is simply $a_{37} \cdot a_{38} \cdot a_{39} \cdot a_{40}$, which is equal to

$$
a_{1} a_{2} a_{3} a_{4}=\frac{a_{2}^{2}}{a_{1}}.
$$

Since $80=13 \cdot 6+2$, the product of the first 80 elements is simply $a_{1} \cdot a_{2}$. Thus, we obtain the following system of equations for the first two terms of the sequence:

$$
\frac{a_{2}^{2}}{a_{1}}=8, \quad a_{1} a_{2}=8
$$

Multiplying the two equations, we get $a_{2}^{3}=64$, so $a_{2}=4$, and thus $a_{1}=2$. Therefore, the first element of the sequence is 2, and the second is 4.