It depends on how you see undirected edges in presence of directed edges. Depending on your need, you can have your own definition of 'strongly connected' and define it accordingly.
As far as I know, if one says 'directed graph' then one usually means that all edges are directed. And if a graph is not directed, then it is undirected.
Here, you can also treat undirected edges as 'bi-directed' edges i.e. you can traverse in any direction on these edges. If you see undirected edges this way then yes, you can call a graph which has at least one directed edge, a 'directed graph'.
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).
Best Answer
Let $T$ be a tournament in which reversing arcs $a$ and $b$ results in a tournament that's not strongly connected. Let $T^a$, $T^b$, and $T^{ab}$ be the tournaments with arc $a$, arc $b$, and both arcs reversed, respectively.
In order for $T^{ab}$ not to be strongly connected, there must be a set $S \subseteq V(T)$ such that $T^{ab}$ has no arcs directed from $S$ to $V(T) - S$.
Meanwhile, $T^a$ and $T^b$ are both strongly connected, so in particular they must each have a way to get from $S$ to $V(T)-S$. The only option that $T^a$ has that $T^{ab}$ does not is arc $b$: so arc $b$ must be directed from $S$ to $V(T)-S$ in $T^a$ (and in $T$). Similarly, the only option that $T^b$ has that $T^{ab}$ does not is arc $a$: so arc $a$ must be directed from $S$ to $V(T)-S$ in $T^b$ (and in $T$).
Therefore in $T$, there are exactly two arcs directed from $S$ to $V(T)-S$: arcs $a$ and $b$.