3. How many five-digit natural numbers have exactly two even digits?
3. First, let's assume that the first digit is odd. Then, for the two even digits, we need to choose two positions from $2,3,4,5$ (looking from left to right), which can be done in $\frac{4 \cdot 3}{2}=6$ ways. After choosing which positions will have even digits and which will have odd digits, for each digit we have 5 choices (for even digits, between $0,2,4,6,8$, and for odd digits, between $1,3,5,7,9$). This gives $5^{5}=3125$ numbers for each of the possibilities with fixed positions of even digits, i.e., $6 \cdot 3125=18750$ numbers in total in the case where the first digit is odd.

Now, let's assume that the first digit is even. Then, for the remaining even digit, we need to choose one of the positions $2,3,4,5$, which can be done in 4 ways. After making this choice, for the first digit we have 4 choices (between $2,4,6,8$, i.e., the first digit cannot be 0), and for all other digits, we have 5 choices. This gives $4 \cdot 5^{4}=2500$ numbers for each of the possibilities with a fixed position of the second even digit, i.e., $4 \cdot 2500=10000$ numbers in total in the case where the first digit is even.

Therefore, in total, there are $18750 + 10000 = 28750$ such numbers.