8.2. Find the largest natural number with all distinct digits such that the sum of any two of its digits is a prime number.
Answer: 520.

Solution: If the desired number is at least a four-digit number, then it either has three digits of the same parity or two pairs of digits of the same parity. In each of these cases, we get that two of the sums of the digits are even. Therefore, they must equal 2. The number 2 can be represented as the sum of different digits in only one way: $0+2$. A three-digit number that satisfies the condition is 520. We will show that there cannot be a larger number. A three-digit number has two digits of the same parity, and as we showed above, these digits can only be 0 and 2. The third digit must be odd. The digit 9 does not work: $9+0=9$ - a composite number. The digit 7 does not work: $7+2=9$. But the digit 5 works.

Comment: Correct answer without justification - 2 points. Proven that the number is no more than three digits - 2 points.