5. A ten-digit number, whose digits can only be 2 or 3, and there are no two adjacent 3s. How many such ten-digit numbers are there?

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5.144.

Solution 1: Consider all positive integers with the given property (using a recursive method).

The number of ten-digit numbers starting with 2 is the same as the number of nine-digit numbers;

Ten-digit numbers starting with 3 must start with 32, and their number is the same as the number of eight-digit numbers, i.e., the number of ten-digit numbers is the sum of the number of nine-digit numbers and the number of eight-digit numbers. This pattern is the same as the Fibonacci sequence.

There are 2 one-digit numbers, 3 two-digit numbers, so there are 5 three-digit numbers, 8 four-digit numbers, 13 five-digit numbers, 21 six-digit numbers, 34 seven-digit numbers, 55 eight-digit numbers, 89 nine-digit numbers, and 144 ten-digit numbers.

Solution 2: Since there cannot be more than two consecutive 3s, the ten digits of a ten-digit number can be divided into five pairs, so there can be at most 5 threes. Since there cannot be more than two consecutive 3s, it can be considered that after determining the number of 2s, 3s are inserted among them.
All positive integers with the given property can be divided into:
(1) If it has no digit 3, then there is only one possible number: 2222222222;
(2) If it has 1 three, then there are 9 twos, and there are $\mathrm{C}_{9+1}^{1}$ $=10$ cases;
(3) If it has 2 threes, then there are 8 twos, and there are $\mathrm{C}_{8+1}^{2}$ $=\frac{9!}{2!\times 7!}=36$ cases;
(4) If it has 3 threes, then there are 7 twos, and there are $\mathrm{C}_{7+1}^{3}$ $=\frac{8!}{3!\times 5!}=56$ cases;
(5) If it has 4 threes, then there are 6 twos, and there are $\mathrm{C}_{8+1}^{4}$ $=\frac{7!}{4!\times 3!}=35$ cases;
(6) If it has 5 threes, then there are 5 twos, and there are $\mathrm{C}_{5+1}^{5}$ $=\frac{6!}{5!}=6$ cases.

Therefore, there are $1+10+36+56+35+6=144$ ten-digit numbers.