(1) By means of graph theory, use $8$ vertices to denote $8$ persons. If two persons know each other, we connect them with an edge. With the given condition, there will be a triangle in every induced subgraph with five vertices, while every triangle in the graph belongs to different $\binom{8-3}{2} = \binom{5}{2} = 10$ induced subgraphs with five vertices. We know that there are $3 \times \binom{8}{5} = 3 \times 56 = 168$ edges in total in these triangles, while every edge is computed ten times repeatedly.

Thus every vertex is incident with at least $\frac{2 \times 168}{8 \times 10} > 4$ edges. So there exists one vertex $A$ that is incident with at least five edges.

Suppose the vertex $A$ is adjacent with five vertices $B$, $C$, $D$, $E$, $F$. By the condition, there exists one triangle in five vertices. Without loss of generality, let $\triangle BCD$ denote the triangle. So there exists one edge between any two vertices in the four vertices $A$, $B$, $C$, $D$. So the corresponding four persons of the four vertices know each other.

(2) If there exist three persons (in a group of six) who know one another in a cyclical manner, there may not exist four persons who know each other.

For example, let $8$ vertices denote $8$ persons. If two persons know each other, we join them with an edge. Consider the regular octagon, we link up the $8$ shortest diagonals, as desired.