2. How many three-digit numbers less than 500 do not contain 1?
2. Let's determine which digit can represent the hundreds place of $-2.3$ or 4. Consider numbers of the form $2 x x$. There will be a total of one hundred such numbers. The digit 1 can be in the tens or units place. If 1 is in the tens place ($21 x$), we get the numbers: $210, 211, 212, 213, 214, 215, 216, 217, 218, 219$. There are ten such numbers. In each ten, 1 can be in the units place:

$$
\text { 201, 211, 221, 231, 241, 251, 261, 271, 281, } 291 .
$$

There are also ten such numbers. Since the number 211 appears twice, the total number of numbers containing the digit 1 will be $19(10+9=19)$. Therefore, the number of numbers not containing the digit 1 in three-digit numbers of the form $2 x x$ will be $81(100-19=81)$.

Similarly, considering numbers of the form $3 x x$ and $4 x x$, we can establish that the total number of three-digit numbers less than 500 that do not contain the digit 1 will be $243(81 \times 3=243)$.

Answer: 243 numbers.