1. Determine a pure periodic decimal that is greater than $\frac{1}{4}$ and less than $\frac{1}{3}$, and the sum of the digits in the period is 12 more than the square of the number of those digits.
Solution. Let $n$ denote the number of digits in the period of the number $x$. From $\frac{1}{4}<x<\frac{1}{3}$, it follows that the first digit of the period is 2 or 3. If it is 3, then the next digit cannot be greater than 3. However, according to the conditions of the problem, the following inequality must hold:

$$
3+3+9(n-2) \geq n^{2}+12
$$

which is easily seen to be impossible. Therefore, the first digit of the period of the number $x$ is equal to 2. Now, similarly as above, the following inequality must hold:

$$
2+9(n-1) \geq n^{2}+12
$$

The only natural numbers for which this inequality holds are $n=4$ and $n=5$.

In the case $n=4$, we need to determine three decimal digits whose sum is equal to $4^{2}+12-2=26$. This can only be the digits 9, 9, and 8 (in any order). In the case $n=5$, we need to determine four decimal digits with a sum of $5^{2}+12-2=35$. These are the digits 9, 9, 9, and 8 (in any order). The numbers

0,(2998); 0,(2989); 0,(2899); 0,(29998); 0,(2,9989); 0,(29899), 0,(28999)

satisfy the conditions of the problem.