Let $S:=\mathcal{S}$ and $a\cdot b:=\langle a,b \rangle$. Without loss of generality $x=0$ (or replace $S$ by $S-x$). That $(s^*,i*)$ is a solution to the max-min optimization problem means the following: $\forall s\in S$ $\exists i_s\in[m]:=\{1,\dots,m\}$ $\forall i\in[m]$ \begin{equation} v_i\cdot s\ge v_{i_s}\cdot s \tag{1} \end{equation} and $\forall s\in S$ \begin{equation} v_{i^*}\cdot s^*\ge v_{i_s}\cdot s, \tag{2} \end{equation} where $i^*:=i_{s^*}$. So, (1) immediately implies \begin{equation} v_i\cdot s^*\ge v_{i_{s^*}}\cdot s^*= v_{i^*}\cdot s^*, \end{equation} which answers your Question 1 affirmatively.

As for Question 2, selecting $S$ and $v_i$'s at random easily provides a counterexample, resulting in the answer "No" to Question 2. E.g., let $n=2$, $x=0\in\mathbb R^2$, $S:=\{(z,w)\in\mathbb R^2\colon z\ge-10,w\ge-10,z+w\le20\}$, $v_1=(3,9)$, $v_2=(8,2)$. Then \begin{equation} \max_{s\in S}\min_{j\in[m]} v_j\cdot s=110<220=\min_{j\in[m]}\max_{s\in S} v_j\cdot s. \end{equation}

Let $S:=\mathcal{S}$ and $a\cdot b:=\langle a,b \rangle$. Without loss of generality $x=0$ (or replace $S$ by $S-x$). That $(s^*,i^*)$ is a solution to the max-min optimization problem means the following: $\forall s\in S$ $\exists i_s\in[m]:=\{1,\dots,m\}$ $\forall i\in[m]$ \begin{equation} v_i\cdot s\ge v_{i_s}\cdot s \tag{1} \end{equation} and $\forall s\in S$ \begin{equation} v_{i^*}\cdot s^*\ge v_{i_s}\cdot s, \tag{2} \end{equation} where $i^*:=i_{s^*}$. So, (1) immediately implies \begin{equation} v_i\cdot s^*\ge v_{i_{s^*}}\cdot s^*= v_{i^*}\cdot s^*, \end{equation} which answers your Question 1 affirmatively.

As for Question 2, selecting $S$ and $v_i$'s at random easily provides a counterexample, resulting in the answer "No" to Question 2. E.g., let $n=2$, $x=0\in\mathbb R^2$, $S:=\{(z,w)\in\mathbb R^2\colon z\ge-10,w\ge-10,z+w\le20\}$, $v_1=(3,9)$, $v_2=(8,2)$. Then \begin{equation} \max_{s\in S}\min_{j\in[m]} v_j\cdot s=110<220=\min_{j\in[m]}\max_{s\in S} v_j\cdot s. \end{equation}

Let $S:=\mathcal{S}$ and $a\cdot b:=\langle a,b \rangle$. Without loss of generality $x=0$ (or replace $S$ by $S-x$). That $(s^*,i*)$ is a solution to the max-min optimization problem means the following: $\forall s\in S$ $\exists i_s\in[m]:=\{1,\dots,m\}$ $\forall i\in[m]$ \begin{equation} v_i\cdot s\ge v_{i_s}\cdot s \tag{1} \end{equation} and $\forall s\in S$ \begin{equation} v_{i^*}\cdot s^*\ge v_{i_s}\cdot s, \tag{2} \end{equation} where $i^*:=i_{s^*}$. So, (1) immediately implies \begin{equation} v_i\cdot s^*\ge v_{i_{s^*}}\cdot s^*= v_{i^*}\cdot s^*, \end{equation} which answers your Question 1 affirmatively.

As for Question 2, almost any $S$ and $v_i$'s should provide a counterexample, resulting in the answer "(Almost Always) No" to Question 2.

Let $S:=\mathcal{S}$ and $a\cdot b:=\langle a,b \rangle$. Without loss of generality $x=0$ (or replace $S$ by $S-x$). That $(s^*,i*)$ is a solution to the max-min optimization problem means the following: $\forall s\in S$ $\exists i_s\in[m]:=\{1,\dots,m\}$ $\forall i\in[m]$ \begin{equation} v_i\cdot s\ge v_{i_s}\cdot s \tag{1} \end{equation} and $\forall s\in S$ \begin{equation} v_{i^*}\cdot s^*\ge v_{i_s}\cdot s, \tag{2} \end{equation} where $i^*:=i_{s^*}$. So, (1) immediately implies \begin{equation} v_i\cdot s^*\ge v_{i_{s^*}}\cdot s^*= v_{i^*}\cdot s^*, \end{equation} which answers your Question 1 affirmatively.

As for Question 2, selecting $S$ and $v_i$'s at random easily provides a counterexample, resulting in the answer "No" to Question 2. E.g., let $n=2$, $x=0\in\mathbb R^2$, $S:=\{(z,w)\in\mathbb R^2\colon z\ge-10,w\ge-10,z+w\le20\}$, $v_1=(3,9)$, $v_2=(8,2)$. Then \begin{equation} \max_{s\in S}\min_{j\in[m]} v_j\cdot s=110<220=\min_{j\in[m]}\max_{s\in S} v_j\cdot s. \end{equation}