2. How many increasing finite sequences of natural numbers exist whose first element is equal to 1, the last element is equal to 25, and any two consecutive members differ by 2 or 3?
2. First solution. Let $x$ be the number of pairs of consecutive terms in the sequence that differ by 2, and $y$ be the number of pairs that differ by 3. Clearly, it must hold that $2x + 3y = 25 - 1 = 24$, and from this it follows that $3 \mid x$, and based on this, we can easily list all possibilities:

$$
(x, y) \in \{(12,0),(9,2),(6,4),(3,6),(0,8)\}
$$

For each such pair $(x, y)$, we will count how many of the desired sequences correspond to this pair. In a sequence corresponding to a pair $(x, y)$, there are a total of $x + y$ differences between consecutive terms, of which $x$ must be equal to 2, and the remaining $y$ must be equal to 3; thus, there are $\binom{x+y}{y}$ such sequences. Therefore, the desired number is

$$
\binom{12}{12} + \binom{11}{9} + \binom{10}{6} + \binom{9}{3} + \binom{8}{0} = 1 + 55 + 210 + 84 + 1 = 351
$$

Second solution. Let $A_{n}$ be the number of increasing finite sequences whose first element is 1, the last is $n$, and any two consecutive terms differ by 2 or 3. By separating cases based on whether the last difference is 2 or 3, we obtain the recurrence relation $A_{n} = A_{n-2} + A_{n-3}$. Given the initial conditions $A_{1} = 1, A_{2} = 0, A_{3} = 1$, we can easily compute the first 25 elements of the sequence:

$$
1,0,1,1,1,2,2,3,4,5,7,9,12,16,21,28,37,49,65,86,114,151,200,265,351
$$

so the solution to the problem is $A_{25} = 351$.