2. A novice is shooting baskets. For each hit, she gets 1 point, and for each miss, $k$ points are deducted, where $k$ is a fixed natural number. (At the beginning, she has 0 points, and the number of points earned at any point during the training can be negative.) At the end of the training, the novice had $k$ points. If it is known that the novice hit more than $75 \%$ but less than $80 \%$ of the total number of shots, determine $k$.
2. Let $d$ be the number of successful shots, and $n$ be the number of missed shots during training. Then we have $d-k n=k$, and according to the problem statement,

$$
\frac{3}{4} < \frac{d}{d+n} < \frac{5}{6}
$$

The left inequality gives $3(d+n) < 4d$, which simplifies to $3n < d$. The right inequality gives $4(d+n) < 5d$, which simplifies to $4n < d$. Combining these, we have $3n < d < 4n$.

Substituting $d = k + kn$ into the inequalities, we get:

$$
3n < k + kn < 4n
$$

This can be rewritten as:

$$
3n < k(1 + n) < 4n
$$

Dividing by $1 + n$, we get:

$$
\frac{3n}{1+n} < k < \frac{4n}{1+n}
$$

For $n = 0$, the inequalities become $0 < k < 0$, which is impossible. For $n = 1$, the inequalities become $1.5 < k < 2$, which is also impossible since $k$ must be an integer. For $n = 2$, the inequalities become $2 < k < 2.67$, which is impossible since $k$ must be an integer. For $n \geq 3$, the left inequality gives $k > \frac{3n}{1+n} > \frac{3n}{2n} = 1.5$, and the right inequality gives $k < \frac{4n}{1+n} < \frac{4n}{n} = 4$. Therefore, $2 < k < 4$, which means $k = 3$.

Thus, the only possible value for $k$ is $3$.