After reading relep's answer I revisited the problem and came up with a different fairly simple proof for Question 1, but before I get to that, I recently found that this problem has a long history already (the key phrase that led me to these references was "non-critical vertex"). It turns out that in 1966 Korvin asked exactly my Question #1 (G. Korvin, *Some combinatorial problems on complete directed graphs*) Then in 1972, Rao and Rao solved the problem raised by Korvin (see Section 2 of S. B. Rao, A. R. Rao, *The number of cut vertices and cut arcs in a strong directed graph*). According to D. Meierling and L. Volkmann, *On the number of nonseparating vertices in strongly connected local tournaments*, Las Vergnas answered Question #2 in the affirmative in 1975, (M. Las Vergnas, *Sur le nombre de circuits dans un tournoi fortement connexe*), but in 1989 Thomassen also answered Question #2 in the affirmative and while he cites the Las Vergnas paper, he doesn't attribute that result to Las Vergnas (C. Thomassen, *Whitney’s 2-switching theorem, cycle spaces, and arc mappings of directed graphs*). Since the Las Vergnas paper is in French and I don't even have easy access to it, I won't be getting to the bottom of that just yet.

As for Rao and Rao's solution to Question #1, there is actually something about their proof that I don't understand. They prove it in the more general setting of semi-complete digraphs and their proof easily shows that if $T$ has some vertex $v$ such that $T-v$ is strongly connected, then $T$ has at least two such vertices. But there seems to be an oversight in the proof that $T$ has at least one such vertex. It has to do with the sentence "Since $G$ is maximal, $x$ is a non-cut vertex of $G_1$." While adding the edge $(y,z)$ "converts some cut vertex into a non-cut vertex," I don't see why that vertex must be $x$.

Finally, here is my different proof of Question 1. Let's call a vertex $v$ in a strongly connected digraph *non-critical* if $T-v$ is strongly connected and *critical* otherwise.

**Lemma 1** If $T$ is a tournament on $n\geq 4$ vertices with diameter $2$, then $T$ has at least $n-2$ non-critical vertices.

*Proof.* If every vertex is non-critical we are done, so suppose $v$ is critical vertex. This implies that $T-v$ has at least two strongly connected components. Suppose $T-v$ has $k\geq 3$ strongly connected components, $T_1$, $\dots$, $T_k$ with all edges oriented from $T_i$ to $T_j$ for all $i<j$. For all $i\in [3]$, let $x_i\in T_i$. Since $T$ has diameter 2, there is a path of length 2 from $x_3$ to $x_2$ which must go through $v$ and there is a path of length 2 from $x_2$ to $x_1$ which also must go through $v$, but this is impossible as it would mean that $(x_2,v)$ and $(v, x_2)$ are edges. So $T-v$ has exactly two strongly connected components $T_1$ and $T_2$ with all edges going from $T_1$ to $T_2$. Furthermore, all edges from $v$ to $T_1$ and all edges from $T_2$ to $v$. If $T_i$ has at least two vertices, then every vertex in $T_i$ is non-critical. Since $n\geq 4$, at least one of $T_1$ or $T_2$ has at least two vertices.$\square$

Say that an $x,y$-path $P$ in a digraph $D$ is a *maximal short path* if $P$ is the shortest path from $x$ to $y$ and there is no other shortest path between two vertices containing $P$ as a segment

**Lemma 2** Let $T$ be a tournament on $n\geq 4$ vertices. If $P$ is a maximal short path of length at least 3 with endpoints $x$ and $y$, then $x$ and $y$ are non-critical vertices.

*Proof.* Let $x,y\in V(T)$ and $P$ as given in the statement. Suppose for contradiction that, say $T-y$ is critical and let $u$ and $v$ be vertices in $T-y$ such that every $u,v$-path goes through $y$. Note that $u$ and $v$ can be chosen so that the distance between $u$ and $v$ is exactly 2. There must be a path from $x$ to $v$ in $T-y$, otherwise the shortest path from $x$ to $v$ would contain $P$. This implies that $(u,x)$ is not an edge; i.e. $(x,u)$ is an edge. But then the distance from $x$ to $y$ is at most 2, contradicting the assumption.$\square$

**Theorem 1** If $T$ is a tournament on $n\geq 4$ vertices, then $T$ has at least two non-critical vertices.

*Proof.* Let $d$ be the diameter of $T$. If $d=2$, we are done by Lemma 1 and if $d=n-1$, then $T\simeq T^*_n$; so suppose $3\leq d\leq n-2$. Let $x$ and $y$ be vertices which witness the fact that the diameter is $d$ and let $P$ be the path from $x$ to $y$. By Lemma 2, $x$ and $y$ are non-critical. $\square$

I feel that this proof should be able to also answer Question 2, but I don't see how at the moment. I originally thought I saw a way forward, but my initial idea didn't work out.