1. Consider the graph $G=(V,M)$ of vertices $V = \{1,2,\ldots,n\}$ and edges $M$. It is trivial to see that $G$ contains no triangle, since if we suppose there exist $1\leq i<j<k\leq n$ such that $(i,j), (i,k) \in M$, then $(j,k) \not \in M$. Then Turán's theorem yields $|M| \leq \lfloor n/2 \rfloor \cdot \lceil n/2 \rceil = \lfloor n^2/4 \rfloor$, with a model for the extremal case of equality given by the complete bipartite graph $V = A \cup B$, where $A = \{1,\ldots,\lfloor n/2 \rfloor\}$ and $B = \{\lceil n/2 \rceil,\ldots,n\}$. 2. Consider the digraph $\vec{G} = (V,M)$. There exists no directed path of length $2$, therefore, after eliminating the isolated vertices, the graph becomes bipartite, with $A \cap B = \emptyset$, $A \cup B \subseteq V$, where $A = \pi_1(M)$ and $B = \pi_2(M)$, for the projections $\pi_1, \pi_2 : M \to V$, given by $\pi_1(j,k) = j$ and $\pi_2(j,k) = k$. Denote $a = |A|$ and $b=|B|$, so $a+b\leq n$. Then $|M| \leq ab$ and it is trivial to see that the maximum of $ab$ is obtained when $a,b$ are $\lfloor n/2 \rfloor, \lceil n/2 \rceil$ (in any order), with the conclusion of above (in fact, part of Turán's theorem proof).