Example 1 Let $A=\{x \mid x \geqslant 10, x \in \mathbf{N}\}, B \subseteq A$, and the elements in $B$ satisfy:
(i) The digits of any element are all different;
(ii) The sum of any two digits of any element is not equal to 9.
(1) Find the number of two-digit and three-digit numbers in $B$;
(2) Does there exist a five-digit or six-digit number in $B$;
(3) If the elements in $B$ are arranged in ascending order, find the 1081st element.
(1) For two-digit numbers, the digit in the tens place can be $1,2, \cdots, 9$; the digit in the units place, since it cannot be the same as the digit in the tens place and the sum of the two digits cannot be 9, has 8 possible choices for each digit in the tens place.
Therefore, the total number of two-digit numbers that meet the criteria is $9 \times 8=72$.
For three-digit numbers, first consider the digit in the hundreds place, which can be $1,2, \cdots, 9$. Next, consider the digit in the tens place, which cannot be the same as the digit in the hundreds place and the sum of the two digits cannot be 9, so there are 8 possible choices. Finally, consider the digit in the units place, which cannot be the same as the digits in the hundreds and tens places, and the sum with the digits in the hundreds and tens places cannot be 9, so there are 6 possible choices.
Therefore, the number of three-digit numbers that meet the criteria is
$$
9 \times 8 \times 6=432 \text{ (numbers). }
$$
(2) There exists a five-digit number that meets the criteria (e.g., 12340).

There does not exist a six-digit number that meets the criteria. The reason is as follows.
Following the method in (1), there are 9 possible choices for the hundred-thousands place, 8 for the ten-thousands place, 6 for the thousands place, 4 for the hundreds place, 2 for the tens place, and 0 for the units place, which is a contradiction.
(3) From (1), we know that there are 72 two-digit numbers and 432 three-digit numbers that meet the criteria.
There are also
$$
9 \times 8 \times 6 \times 4=1728
$$

numbers that meet the criteria among four-digit numbers, and among these, there are $8 \times 6 \times 4=192$ numbers with 1 in the thousands place, and 192 numbers each with 2 or 3 in the thousands place.

Since $1081-(72+432+192 \times 3)=1$, the number that meets the criteria is the smallest number with 4 in the thousands place, i.e., 4012 is the 1081st element of $B$.