25. There are 8 items with different masses, each of which is an integer number of grams and does not exceed 15 grams. Xiao Ping wants to find the item with the greatest mass using the fewest number of weighings on a balance. He used the following method:
(1) Divide the 8 items into 2 groups, each with 4 items, and compare the masses of these 2 groups;
(2) Take the 4 items from the heavier group and divide them into 2 groups, each with 2 items, and compare their masses;
(3) Take the 2 items from the heavier group and divide them into 1 item each, and take the heavier one.
Xiao Ping did not get a balanced result on the balance in all 3 weighings, and finally obtained one item.
However, the item he found is actually the 5th heaviest among the 8 items.
Question: What is the mass of the item Xiao Ping found? And what is the mass of the second heaviest item?
25. Let the weights of these 8 items, from heaviest to lightest, be arranged as follows:
$$
15 \geqslant a_{1}>a_{2}>a_{3}>a_{4}>a_{5}>a_{6}>a_{7}>a_{8} \geqslant 1 .
$$

Xiao Ping found the item with weight $a_{5}$, and the second lightest item has weight $a_{7}$.
Since adding a lighter item to $a_{5}$ cannot exceed the sum of the weights of two items heavier than $a_{5}$, the first step must eliminate 3 items heavier than $a_{5}$.
Thus, there are the following four possibilities:
First case $\left\{\begin{array}{l}a_{4}+a_{5}+a_{5}+a_{7}>a_{1}+a_{2}+a_{3}+a_{8}, \\ a_{5}+a_{6}>a_{4}+a_{7}, \\ a_{5}>a_{6},\end{array}\right.$
Second case $\left\{\begin{array}{l}a_{3}+a_{5}+a_{6}+a_{7}>a_{1}+a_{2}+a_{4}+a_{8}, \\ a_{5}+a_{6}>a_{3}+a_{7}, \\ a_{5}>a_{6},\end{array}\right.$
Third case $\left\{\begin{array}{l}a_{2}+a_{5}+a_{6}+a_{7}>a_{1}+a_{3}+a_{4}+a_{8}, \\ a_{5}+a_{6}>a_{2}+a_{7}, \\ a_{5}>a_{6},\end{array}\right.$
Fourth case $\left\{\begin{array}{l}a_{1}+a_{5}+a_{6}+a_{7}>a_{2}+a_{3}+a_{4}+a_{8} . \\ a_{5}+a_{6}>a_{1}+a_{7}, \\ a_{5}>a_{6} .\end{array}\right.$
First, consider the first case. According to (1), $a_{4}$ is at least 3 grams lighter than $a_{1}$, $a_{5}$ is at least 3 grams lighter than $a_{2}$, and $a_{6}$ is at least 3 grams lighter than $a_{3}$. Thus, $a_{7}$ is at least 10 grams heavier than $a_{8}$. According to (2), $a_{3}$ is at least 1 gram lighter than $a_{4}$, so $a_{6}$ is at least 18 grams heavier than $a_{7}$, which contradicts the given conditions. Therefore, the first case is impossible.
Using the same reasoning, it can be shown that the second and third cases are also impossible.
Finally, consider the fourth case. $a_{1}$ is at least 1 gram heavier than $a_{2}$; $a_{5}$ is at least 2 grams lighter than $a_{3}$, and $a_{6}$ is at least 2 grams lighter than $a_{4}$, so $a_{7}$ is at least 4 grams heavier than $a_{8}$. According to (4), $a_{5}$ is at least 4 grams lighter than $a_{1}$, so $a_{6}$ is at least 5 grams heavier than $a_{7}$. Thus, the weights of these 8 items are:
$a_{1}=15$ grams, $a_{2}=14$ grams, $a_{3}=13$ grams, $a_{4}=12$ grams,
$a_{5}=11$ grams, $a_{6}=10$ grams, $a_{7}=5$ grams, $a_{8}=1$ gram,
Therefore, the item Xiao Ping found weighs 11 grams, and the second lightest item weighs 5 grams.