2. First method: Find the digits $a, b, c$ such that the number $\overline{387 a b c}$ is divisible by 5, 6, and 7. Any even number divisible by 5 ends in 0, so $c=0$. For divisibility by 3, it is necessary that $3+8+7+a+b=18+a+b$ is divisible by 3. Since 18 is divisible by 3, $a+b$ must be divisible by 3. Let's take the smallest number divisible by 6 and 5, which is 387030. Direct verification shows that this number is also divisible by 7, and therefore is the desired number. If we add $210=5 \cdot 6 \cdot 7$ to this number, its divisibility by 5, 6, and 7 will not change. Therefore, 387240, 387450, 387660, 387870 are also solutions.

Second method: A number divisible by 5, 6, and 7 is divisible by $210=5 \cdot 6 \cdot 7$ and has the form $210k$. We need to find $k$ such that $387000 \leq 210k \leq 387999$. We get $k=1843, \ldots 1847$.
Answer: 387030, 387240, 387450, 387660, 387870.

Evaluation recommendations: Students may find only one or a few solutions. If only one solution is found, the task is scored 3 points; if several but not all solutions are found - 5 points. Full solution - 7 points.