1. According to the condition, these numbers are written only with the digits $0,1,2,6,8$. Then, three-digit numbers that are multiples of 4 can end with exactly 10 variants: $00,08,12,16,20,28,60,68,80,88$. In each of these 10 variants, the first place can be occupied by any of the 4 digits $1,2,6,8$. Additionally, the last 8 variants give two-digit numbers, and the second one gives a single-digit number. Therefore, the total number of such numbers is: $10 \cdot 4+8+1=49$.
# Answer: 49.

Answer to option: 1-2: 53.

## Solutions for option 1-2.

According to the condition, these numbers are written only with the digits $0,2,3,4,5,7$. Then, three-digit numbers that are multiples of 4 can end with exactly 9 variants: $00,04,20,24,32,40,44,52,72$. In this case, the first place in each of these 9 variants can be occupied by any of the 5 digits $2,3,4,5,7$. Additionally, the last 7 variants give two-digit numbers, and the second one gives a single-digit number. Therefore, the total number of such numbers is: $9 \cdot 5 + 7 + 1 = 53$.