7.14 In the sequence $19752 \cdots$, starting from the 5th digit, each digit equals the unit digit of the sum of the previous 4 digits. Try to answer the following questions:
(1) Will the digit groups 1234 and 3269 appear in the sequence?
(2) Will the digit group 1975 appear again in the sequence?
(3) Will the digit group 8197 appear in the sequence?
[Solution] (1) We replace all even digits in the sequence with 0, and all odd digits with 1, then the first five digits become 11110.

According to the rule of the sequence, the 5th digit equals the unit digit of the sum of the previous 4 digits, so this "1,0" sequence will periodically repeat 11110.
The "1,0" sequences corresponding to 1234 and 3269 are 1010 and 1001, respectively.
Therefore, they cannot appear in the given sequence.
(2) Since there are only 10 digits, only a finite number of 4-digit numbers can be formed, while the entire sequence is infinite. Therefore, in the entire sequence, we can definitely find two identical 4-digit numbers $A$: i.e.,
$$
1975 \cdots A \cdots A \cdots
$$

Next, we prove that the number 1975 must exist between these two identical 4-digit numbers.
For this, we can extend the given sequence to the left. That is, once we know 4 consecutive digits, we can not only determine the digit that follows them but also the digit that precedes them.

Thus, when we know a segment of the sequence $A \cdots A$, we can obtain an infinite sequence in both directions. If 1975 does not appear between the two $A$s, then it will not appear anywhere in the entire sequence, which contradicts the fact that the given sequence contains 1975.
(3) From (2), the digit before 1975 is 8, which can be derived from
$$
8+1+9+7=25
$$

Thus, we can find the digit group 8197 in the infinite sequence in both directions, and therefore it can also be found in the original sequence.