【Solution】Solution: (1) When the unit is 0:
We need to choose one more from the remaining 5 numbers, the number of 0s can be two or three,
When there are two 0s, there are 4 arrangement ways, when there are three 0s, there are 6 arrangement ways,
So there are in total:
$$
5 \times(4+6)=50 \text { (numbers) }
$$

Among them, the ones with the highest digit being 0 are:
$$
5 \times(1+3)=20 \text { (numbers) }
$$

The ones that meet the conditions are:
$$
50-20=30 \text { (numbers) }
$$
(2) When the unit is not 0, it can be 2 or 4, two options, we need to choose one more from the remaining 5 numbers, when 2 or 4 has two, there are 4 arrangement ways, when 2 or 4 has three, there are 6 arrangement ways, so there are in total:
$$
2 \times 5 \times(4+6)=100 \text { (numbers) }
$$

Among them, the ones with the highest digit being 0 are:
$$
2 \times(3+3)=12 \text { (numbers) }
$$

So the ones that meet the conditions are:
$$
100-12=88 \text { (numbers) }
$$

Therefore, in total:
$$
30+88=118 \text { (numbers) }
$$

Answer: There are 118 five-digit numbers that meet the conditions.
16. (15 points) Use two of the numbers $0, 1, 2, 3, 4, 5$ to form a five-digit even number, where one number appears 2 times and the other number appears 3 times. How many five-digit numbers satisfy this condition?