Four. (50 points) There are 8 items with different weights, each weighing an integer number of grams and not exceeding 15 grams. Xiao Ping wants to find the heaviest item with the fewest number of weighings using a balance scale. He used the following method:
(1) Divide the 8 items into 2 groups, 4 items in each, and compare the weights of these 2 groups;
(2) Take the heavier 4 items from the 2 groups and divide them into 2 groups, 2 items in each, and compare their weights;
(3) Take the heavier 2 items from the 2 groups and divide them into 1 item each, and take the heavier 1 item.
Xiao Ping did not get a balanced result in 3 weighings and finally obtained one item. However, the item he found is actually the 5th heaviest among the 8 items. Question: How much does the item Xiao Ping found weigh? How much does the second lightest item weigh?
Let the weights of the 8 items, from heaviest to lightest, be arranged as follows:
$15 \geqslant a_{1}>a_{2}>a_{3}>a_{4}>a_{5}>a_{6}>a_{7}>a_{8} \geqslant 1$,
The item found by Xiaoping weighs $a_{5}$, and the second lightest item weighs $a_{7}$.
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 remove 3 items heavier than $a_{5}$. This leads to the following four possibilities:
First possibility $\left\{\begin{array}{l}a_{4}+a_{5}+a_{6}+a_{7}>a_{1}+a_{2}+a_{3}+a_{8}, \\ a_{5}+a_{6}>a_{4}+a_{7}, \\ a_{5}>a_{6},\end{array}\right.$
Second possibility $\left\{\begin{array}{l}a_{3}+a_{5}+a_{6}+a_{7}>a_{1}+a_{2}+a_{4}+a_{8}, \\ a_{5}+a_{6}>a_{3}+a_{7}, \\ a_{5}>a_{6},\end{array}\right.$
Third possibility $\left\{\begin{array}{l}a_{2}+a_{5}+a_{6}+a_{7}>a_{1}+a_{3}+a_{4}+a_{8}, \\ a_{5}+a_{6}>a_{2}+a_{7} \\ a_{5}>a_{6},\end{array}\right.$
Fourth possibility $\left\{\begin{array}{l}a_{1}+a_{5}+a_{6}+a_{7}>a_{2}+a_{3}+a_{4}+a_{8}, \\ a_{5}+a_{6}>a_{1}+a_{7}, \\ a_{5}>a_{6} .\end{array}\right.$
First, consider the first possibility. 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 $a_{3}$. Therefore, $a_{7}$ is at least 10 grams heavier than $a_{8}$. According to (2), $a_{5}$ is at least 1 gram lighter than $a_{4}$, so $a_{6}$ is at least 2 grams heavier than $a_{7}$. Combining (1) and (2), $a_{6}$ is at least 12 grams heavier than $a_{8}$. Considering the extreme case where $a_{8}=1$, we get $a_{1}>15$, which contradicts the given condition. Therefore, the first possibility cannot occur.
Using the same reasoning, the second and third possibilities can also be shown to be impossible.
Finally, consider the fourth possibility.
Since $a_{1}$ is at least 1 gram heavier than $a_{2}$, let $a_{1}-a_{2}=k$, where $1 \leqslant k \leqslant 8, k \in \mathbf{N}$.
From $a_{1}+a_{5}+a_{6}+a_{7}>a_{2}+a_{3}+a_{4}+a_{8}$, we get: $\left(a_{1}-a_{2}\right)+\left(a_{7}-a_{8}\right)>\left(a_{3}-a_{5}\right)+\left(a_{4}-a_{6}\right) \geqslant 4$.
Since $a_{1}-a_{2}=k$, we have $a_{7}-a_{8}>4-\left(a_{1}-a_{2}\right)=4-k$, which means $a_{7}-a_{8} \geqslant 5-k$.
From (4), $a_{5}+a_{6}>a_{1}+a_{7}$, we get
$a_{6}-a_{7}>a_{1}-a_{5}=\left(a_{1}-a_{2}\right)+\left(a_{2}-a_{3}\right)+\left(a_{3}-a_{4}\right)+\left(a_{4}-a_{5}\right)=k+3$, which means $a_{6}-a_{7} \geqslant k+4$.
Therefore, $a_{6}-a_{8}=\left(a_{6}-a_{7}\right)+\left(a_{7}-a_{8}\right) \geqslant(5-k)+(k+4)=9$.
This means $a_{6}$ is at least 9 grams heavier than $a_{8}$. Thus, the weights of the 8 items are:
$$
a_{1}=15 \text { grams, } a_{2}=14 \text { grams, } a_{3}=13 \text { grams, } a_{4}=12 \text { grams, }
$$
$a_{5}=11$ grams, $a_{6}=10$ grams, $a_{7}=5$ grams, $a_{8}=1$ gram.
Therefore, the item found by Xiaoping weighs 11 grams, and the second lightest item weighs 5 grams.