The base case holds. Take a king in the tournament. Because the graph is strongly connected, $|N^+(v)|,|N^-(v)| \neq \emptyset$. Therefore, both $|N^+(v)|$ and $|N^-(v)|$ have kings. Let the king of $|N^+(v)|$ be $k_+$. Clearly, it has to reach $v$ with a path of length $2$. Thus, we have a cycle of length three.
Now suppose that $G$ has a cycle $C$ of length $c$. There are two cases:
Easy case: There is some $v \in G -C$ s.t. $v$ loses to a $x \in C$ and beats a vertex $y \in C$. Then we splice $v$ into $C$ by $xvy$, creating a cycle of length $c+1$.
Hard case:
Suppose that no such $v$ exists. Then we split the vertices of $G - C$ into two disjoint sets: $S$, set of vertices that only lose to vertices in $C$, and $P$, the set of vertices that only beat vertices in $C$. Neither of these sets can be nonempty, or the graph is not strongly connected.
There exists $s \in S$ and $p \in P$ s.t. $s$ beats $p$, otherwise the graph cannot be strongly connected.
Take the vertices $c_1, c_2,c_3 \in C$ s.t. $c_3$ follows $c_2$, which follows $c_1$ in $C$. By definition, $c_1$ beats $s$, $s$ beats $p$, and $p$ beats $c_3$. Splice $s$ and $p$ into $C$ by $c_1spc_3$. By skipping $c_2$, we create a cycle of length $c+1$.
Therefore, we have shown by induction that any strongly connected tournament contains cycles of all lengths $3,...,n$, and thus must contain a Hamiltonian cycle.
For more information, you can check out the paper where this result was first published.
This can be proven using strong induction. Let $n$ be the number of vertices. When $n \le 2$, a hamiltonian path clearly exists. Now, for any given $n > 2$, pick any arbitrary vertex $v$. Partition all other vertices other than $v$ into the sets $V_{out}$ and $V_{in}$ -- $V_{out}$ contains all other vertices $u$ such that the edge $(v, u)$ exists, while $V_{in}$ contains all other vertices $u$ such that the edge $(u, v)$ exists.
Clearly $|V_{out}| < n$ and $|V_{in}| < n$, so by the inductive hypothesis, there is a hamiltonian path in both sets. Let $H_{out}$ be any hamiltonian path in $V_{out}$ and $H_{in}$ be any hamiltonian path in $V_{in}$. You can then form a hamiltonian path for all vertices by concatenating $H_{in}$, $v$, and $H_{out}$.
Best Answer
Let's call your tournament $T$, and write $a \rightarrow b$ when there is an arc from $a$ to $b$. Write $a \rightarrow B$ when $B$ is a vertex set and $a$ has an arc to all members of $B$. Finally write $B \rightarrow a$ when all members of a set $B$ have an arc to $a$.
1) is straightforward enough. If $T$ has a cycle of length 3, then it can't be transitive by definition. Conversely, suppose $T$ is not transitive. Then $\forall a,b,c \in V(T) [a \rightarrow b, b \rightarrow c \Rightarrow a \rightarrow c]$ is false. What do you get by negating this expression ?
2) There's probably a better way for the tougher direction, but here goes. One way is to suppose $T$ has no Hamiltonian cycle and get that $T$ is not strongly connected. Let $C = c_1c_2 \ldots c_k c_1$ be a directed cycle of $T$ with the largest number of vertices.
Let $x \in V(T)$ such that $x \notin C$. If $c_i \rightarrow x$ for some $c_i \in C$, then $c_{i + 1} \rightarrow x$ as otherwise, $x \rightarrow c_{i+1}$ and $C' = c_1c_2 \ldots c_i x c_{i + 1} \ldots c_k c_1$ would be a bigger cycle (but $C$ is a largest cycle by assumption). Applying this reasoning inductively, we have that $C \rightarrow x$ when some $c_i \rightarrow x$.
So what do you get for any $x$ outside $C$ ? Either $C \rightarrow x$ or $x \rightarrow C$. Let $X = \{x : C \rightarrow x\}$ and $Y = \{y : y \rightarrow C\}$. Take $x \in X, y \in Y$. If $x \rightarrow y$, then you can extend the $C$ cycle by deviating with $c_ixyc_{i + 1}$. So $y \rightarrow x$, for all $x \in X, y \in Y$. From this, you can deduce that there is no way to go from $x \in X$ to $c_i \in C$, hence $T$ is not strongly connected.
What's left for you to do is to argue that $C$ exists, and handle the cases when $X$ or $Y$ is empty (and tell us why both can't be empty).