(5) In a simple graph with eight vertices, what is the maximum number of edges in a graph that does not contain a quadrilateral? (A simple graph is one where no vertex is connected to itself, and any two vertices are connected by at most one edge.)
Figure 1 is a simple graph with eight vertices and eleven edges, where there is no quadrilateral.

Below we prove that 11 is the required maximum value. For this, it is sufficient to prove that a graph with twelve edges must contain a quadrilateral. First, we point out a few facts:
(1) Suppose point $A$ is connected to points $A_{1}, A_{2}, \cdots, A_{k}$, and $B$ is a vertex different from $A$ (which can be one of the points in $\left\{A_{1}, A_{2}, \cdots, A_{k}\right\}$). If $B$ is connected to two points in $\left\{A_{1}, A_{2}, \cdots, A_{k}\right\}$, then the graph contains a quadrilateral.
(2) In a graph with four vertices, if there are five edges, there must be a quadrilateral.
Let $G$ be a simple graph with eight vertices and twelve edges. We will prove that $G$ must contain a quadrilateral. We use proof by contradiction, i.e., assume $G$ has no quadrilateral. Let $d_{1}, d_{2}, \cdots, d_{8}$ be the degrees of the eight vertices of $G$. Since
$$
\sum_{i=1}^{8} d_{i}=2 \times 12=24, \max \left(d_{1}, d_{2}, \cdots, d_{8}\right) \geqslant 3
$$

We will discuss the following three cases.
i $\max \left(d_{1}, d_{2}, \cdots, d_{8}\right)=3$. In this case, the degree of each vertex in $G$ is 3. Take any vertex $A$, and let the vertices connected to $A$ be $A_{1}, A_{2}, A_{3}$, and the remaining four vertices be $B_{1}, B_{2}, B_{3}, B_{4}$. By (1), the number of edges between $\left\{B_{1}, B_{2}, B_{3}, B_{4}\right\}$ and $\left\{A_{1}, A_{2}, A_{3}\right\}$ is at most four. The number of edges between the vertices in $\left\{A_{1}, A_{2}, A_{3}\right\}$ is at most one, so the number of edges between the vertices in $\left\{B_{1}, B_{2}, B_{3}, B_{4}\right\}$ is at least $12-3-4-1=4$ (by (2), there are also only four edges). We only consider these four vertices and the four edges connecting them. At this time, there must be a vertex with degree 1 (if the degree of these four points is all 2, it will form a quadrilateral). Thus, there is a vertex with degree 3, i.e., there must be a point (let's say $B_{1}$) in $\left\{B_{1}, B_{2}, B_{3}, B_{4}\right\}$ that is connected to the other three points $B_{2}, B_{3}, B_{4}$. And the number of edges between the points in $\left\{B_{1}, B_{2}, B_{3}, B_{4}\right\}$ and the points in $\left\{A_{2}, A_{3}, A_{4}\right\}$ is 4, so $B_{1}$ must be connected to one of the points in $\left\{A_{2}, A_{3}, A_{4}\right\}$. Thus, in the graph $G$, the degree of $B_{1}$ will be 4, which contradicts the assumption.
ii $\max \left(d_{1}, \cdots, d_{8}\right)=4$. Take a vertex $A$ with degree 4, and let the vertices connected to $A$ be $A_{1}, A_{2}, A_{3}, A_{4}$, and the remaining three vertices be $B_{1}, B_{2}, B_{3}$. By (1), the number of edges between the points in $\left\{A_{1}, A_{2}, A_{3}, A_{4}\right\}$ is at most 2, the number of edges between the points in $\left\{B_{1}, B_{2}, B_{3}\right\}$ and the points in $\left\{A_{1}, A_{2}, A_{3}, A_{4}\right\}$ is at most 3, and the number of edges between the points in $\left\{B_{1}, B_{2}, B_{3}\right\}$ is at most 3. Since the total number of edges is 12, the sum of the above edge numbers is $2,3,3$. Therefore, the number of edges between the points in $\left\{A_{1}, A_{2}, A_{3}, A_{4}\right\}$ is 2, let's say $A_{1} A_{2}$ and $A_{3} A_{4}$ (these two edges do not share a common vertex, otherwise a quadrilateral will appear). Each point in $\left\{B_{1}, B_{2}, B_{3}\right\}$ is connected to one of the points in $\left\{A_{1}, A_{2}, A_{3}, A_{4}\right\}$, and $B_{1} B_{2}, B_{1} B_{3}, B_{2} B_{3}$ are all edges in $G$. Since each point in $\left\{B_{1}, B_{2}, B_{3}\right\}$ is connected to one of the points in $\left\{A_{1}, A_{2}, A_{3}, A_{4}\right\}$, at least two points are connected to one of the points in $\left\{A_{1}, A_{2}\right\}$ or $\left\{A_{3}, A_{4}\right\}$, let's say $\left\{B_{1}, B_{2}, B_{3}\right\}$ has two points $B_{1}, B_{2}$ connected to one of the points in $\left\{A_{1}, A_{2}\right\}$. At this time, if $B_{1}, B_{2}$ are both connected to the same point (let's say $A_{1}$), $A_{1} B_{1} B_{3} B_{2}$ is a quadrilateral. If $B_{1}, B_{2}$ are connected to different points in $\left\{A_{1}, A_{2}\right\}$, then because $A_{1} A_{2}, B_{1} B_{2}$ are both edges, there is still a quadrilateral, which contradicts the assumption.
iii $\max \left(d_{1}, \cdots, d_{8}\right)>4$. Take a vertex $A$ with the maximum degree, and let the set of vertices connected to $A$ be $S$, and the remaining vertices be $T$.
$$
|S|=\max \left(d_{1}, \cdots, d_{8}\right)
$$
The number of edges between the points in $T$ and the points in $S$ is at most $|T|$, the number of edges between the points in $T$ is at most $\mathrm{C}_{|T|}^{2}$, and the number of edges between the points in $S$ is at most $\left[\frac{|S|}{2}\right]$. Thus, the number of edges in the graph is at most
$$
|S|+|T|+\mathrm{C}_{|T|}^{2}+\left[\frac{|S|}{2}\right]=7+\mathrm{C}_{|T|}^{2}+\left[\frac{|S|}{2}\right]
$$

When $|S|=5,6,7$, it is impossible for the number of edges to be 12.
In summary, the required maximum value is 11.