2. Find a natural number $N$, such that it is divisible by 5 and 49, and including 1 and $N$, it has a total of 10 divisors.
Let $N=2^{a_{1}} 3^{a_{2}} 5^{a_{3}} 7^{a_{4}} \cdots p_{n}^{a_{n}}\left(a_{i} \in \mathbf{N}\right.$, $i=1,2, \cdots, n)$. Then
$$
\left(a_{1}+1\right)\left(a_{2}+1\right) \cdots\left(a_{n}+1\right)=10 \text {. }
$$

Since $5\left|N, 7^{2}\right| N$, then $a_{3}+1 \geqslant 2, a_{4}+1 \geqslant 3$. Therefore, $a_{1}, a_{2}, a_{5}, a_{6}, \cdots, a_{n}$ must all be 0.
Thus, we solve $N=5 \times 7^{4}$.