2. How many five-digit natural numbers have an even number of even digits?
2. We will first sort the numbers that have

$1^{\circ}$ the first digit even, and then

$2^{\circ}$ those with the first digit odd.

5 points

$1^{\circ}$ The first digit can be one of four digits: $2,4,6,8$.

Of the remaining four digits, one or three must be even.

For these 4 positions, we have 5 possibilities for even and 5 for odd numbers.

In this case, there are

$$
4 \cdot\left[\binom{4}{1}+\binom{4}{3}\right] \cdot 5^{4}=20000
$$

desired numbers.
$2^{\circ}$ Here, the first (odd) digit can be chosen in 5 ways, and of the remaining four, 0, 2, or 4 digits are even. For the remaining 4 positions, we have 5 possibilities for even and 5 for odd numbers. In this case, there are

$$
5 \cdot\left[\binom{4}{0}+\binom{4}{2}+\binom{4}{4}\right] \cdot 5^{4}=25000
$$

desired numbers.

10 points

Therefore, there are a total of $20000+25000=45000$ possibilities.