Second question As shown in Figure 8, in a $7 \times 8$ rectangular chessboard, a chess piece is placed at the center point of each small square. If two chess pieces are in small squares that share an edge or a vertex, then these two chess pieces are said to be "connected". Now, some of the 56 chess pieces are removed so that no 5 remaining chess pieces are in a straight line (horizontally, vertically, or diagonally) in sequence. How many chess pieces must be removed at minimum to meet this requirement? And explain the reasoning.
Solution: The minimum number of chess pieces to be removed is 11.
Define the square at the $x$-th column and $y$-th row as $(x, y)$ $(1 \leqslant x \leqslant 8,1 \leqslant y \leqslant 7, x, y \in \mathbf{Z})$.

A group of 5 consecutive chess pieces is called a "good group". The problem requires that no chess pieces form a "good group".
Let the number of removed chess pieces be $n$.
Lemma 1: There must be a removed chess piece in each row and each column. If no chess piece is removed in a certain row (column), then there will be more than one good group.

Lemma 2: If only 1 chess piece is removed in a certain column, then this chess piece must be in the 3rd, 4th, or 5th row of that column.

Lemma 3: If only 1 chess piece is removed in a certain row, then this chess piece must be in the 4th or 5th column of that row.
Proof of lemmas is omitted.
When $n \leqslant 7$, by the pigeonhole principle, there must be 1 column without any removed chess pieces. By Lemma 1, there must be a good group, so $n \leqslant 7$ does not meet the requirement.

When $n=8$, by Lemma 1, exactly 1 chess piece is removed from each column. By Lemma 2, these 8 chess pieces are all in the 3rd, 4th, or 5th rows. At this point, the 1st, 2nd, 6th, and 7th rows have no removed chess pieces, so there exists a good group. Therefore, $n=8$ does not meet the requirement.

When $n=9$, if 3 or more chess pieces are removed in a certain column, then in the remaining 7 columns, fewer than or equal to 6 chess pieces are removed, which means there must be a column without any removed chess pieces, and thus a good group exists. Therefore, at most 2 chess pieces can be removed in one column. By Lemma 1, a chess piece must be removed from each column. Since $9=1 \times 7+2$, it is only possible that 7 columns each have 1 chess piece removed, and the remaining 1 column has 2 chess pieces removed.

By Lemma 2, the 7 chess pieces in these 7 columns are all in the 3rd, 4th, or 5th rows, and there are 2 remaining chess pieces in the 1st, 2nd, 6th, and 7th rows. Therefore, there must be a row without any removed chess pieces, and thus a good group exists. Therefore, $n=9$ does not meet the requirement.

When $n=10$, if 4 or more chess pieces are removed in a certain column, then in the remaining 7 columns, fewer than or equal to 6 chess pieces are removed, which means there must be a column without any removed chess pieces, and thus a good group exists; if 3 chess pieces are removed in a certain column, then in the remaining 7 columns, 7 chess pieces are removed. By Lemma 1 and Lemma 2, exactly 1 chess piece is removed from each of these 7 columns, and these 7 chess pieces are distributed in the 3rd, 4th, and 5th rows. At this point, the 1st, 2nd, 6th, and 7th rows have no removed chess pieces, and there are 3 remaining chess pieces, which means there must be a row without any removed chess pieces, and thus a good group exists; by Lemma 1 and $10=1 \times 6+2 \times 2$, it is only possible that 6 columns each have 1 chess piece removed, and 2 columns each have 2 chess pieces removed. By Lemma 2, the 6 chess pieces in these 6 columns are distributed in the 3rd, 4th, and 5th rows, and the remaining 4 chess pieces are distributed in the 1st, 2nd, 6th, and 7th rows. By Lemma 1, these 4 chess pieces correspond to the 1st, 2nd, 6th, and 7th rows, one-to-one, without overlap. By Lemma 3, the two columns with 2 chess pieces removed are the 4th and 5th columns. Since the two chess pieces in the same column are distributed in the 1st, 2nd, 6th, and 7th rows, the possible combinations of removed chess pieces are:
$$
\begin{array}{l}
(4,1) \text { and }(4,6),(4,2) \text { and }(4,7), \\
(4,1) \text { and }(4,7),(4,2) \text { and }(4,6),
\end{array}
$$

Among them, $(4,1)$ and $(4,7)$ do not meet the requirement (because $(4,2)$, $(4,3), \cdots,(4,6)$ form a good group).
Similarly, $(5,1)$ and $(5,7)$ do not meet the requirement.
Also, because only 1 chess piece is removed in each row, $(4,2)$ and $(4,6)$ do not meet the requirement.
By symmetry, assume the removed chess pieces are
$(4,1),(4,6),(5,2),(5,7)$.
At this point, as shown in Figure 9 ("○" represents a chess piece, "×" represents a removed chess piece, and blank squares represent undetermined positions).

Since only 1 chess piece is removed in the 3rd column, at most 1 "×" is in $(3,3),(3,5)$, so in $(1,1),(2,2),(3,3),(4,4)$, $(5,5)$ and $(1,7),(2,6),(3,5),(4,4),(5,3)$, there must be 1 good group.

Therefore, $n=10$ does not meet the requirement.

When $n=11$, an example can be constructed as shown in Figure 10.

In summary, the minimum number of chess pieces to be removed is 11.