2. Find the probability that a randomly selected five-digit natural number with non-repeating digits, formed from the digits $1,2,3,4,5,6,7,8$, is divisible by 8 without a remainder. (10 points)
Solution. A natural number $n$ is divisible by 8 without a remainder if the number formed by the last three digits in the number $n$ (in the order of their occurrence) is divisible by 8. If the number formed by the last three digits is not divisible by 8, then the number $n$ is not divisible by 8.

Let's find the number of three-digit natural numbers, composed of non-repeating digits 1, $2,3,4,5,6,7,8$, and divisible by 8. These numbers are also divisible by 4. First, let's find all two-digit numbers, composed of non-repeating digits $1,2,3,4,5,6,7,8$, which are divisible by 4. These numbers are: $12,16,24,28,32,36,48,52,56,64,68,72,76,84$. Among them, 7 numbers are divisible by 8. A three-digit number with the last two digits forming a number divisible by 8 will be divisible by 8 if the first digit of this number is an even number. Indeed, $100 n + 8 k = 8 m, 25 n = 2(m - k), m, n, k \in \mathbb{N}, \Rightarrow n \in \{2,4,6,8\}$. Thus, using two-digit numbers divisible by 8 with the given conditions, we can obtain 18 three-digit numbers divisible by 8. The remaining 7 numbers are divisible by 4 but not by 8, i.e., they can be represented as $4(2 k + 1), k = 1,3,4,6,8,9,10$. Then $100 n + 4(2 k + 1) = 8 m, 25 n + 2 k + 1 = 2 m, m, n, k \in \mathbb{N}, \Rightarrow n \in \{1,3,5,7\}$. Thus, we get another 24 three-digit numbers divisible by 8. In total, we have 42 three-digit natural numbers, composed of non-repeating digits $1,2,3,4,5,6,7,8$, and divisible by 8.

The first two digits of a five-digit number can be chosen from the digits 1, 2, 3, 4, 5, 6, 7, 8, taking into account their differences and differences with the three chosen last digits, i.e., we will have $5 \cdot 4$ ways. The total number of ways to choose a five-digit natural number with non-repeating digits, composed of digits $1,2,3,4,5,6,7,8$, is $A_{8}^{5} = 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4$. The desired probability is $P = \frac{42 \cdot 5 \cdot 4}{8 \cdot 7 \cdot 6 \cdot 5 \cdot 4} = \frac{1}{8}$. Answer: $\frac{1}{8}$.