I am very stuck on this question:
Suppose that $b \in \mathbb{R}^m$, $c \in \mathbb{R}^n$, $A$ is a $m \times n$ real matrix, and all components of $A$, $b$ and $c$ are positive. Consider the two-person zero-sum game in which each player has $m+n+1$ pure strategies and the payoff matrix is \begin{equation} M= \Bigg( \begin{array}{c:c} 0 & A& -b \\ \hdashline -A^{T} & 0 & c \\ \hdashline b^{T} & -c^{T} & 0 \end{array} \Bigg). \end{equation}
There are $3$ parts:
$1.$ Find the equilibrium payoff for player 1 (row player).
$2.$ If both players have the same optimal mixed strategy in the game being \begin{equation} \pi^{T} = (p_1, p_2, \ldots, p_m, q_1, \ldots, q_n, r), \end{equation} prove that $r \neq 0$.
$3.$ Explain how to find from $\pi$ an optimal solution to the linear program \begin{equation} \{ \text{ maximize } c^{T} y \, : \, Ay \leq b, y\geq 0\}. \end{equation}
For 1. Use the following
that you can find in this link or in any textbook.
For 2. Since $π$ is optimal and the game is $0$ sum we know that $π$ guarantees the value of the game (i.e. $0$) for player I against any strategy of player II (we will use also in part 3. below). Assume know that $r=0$, then $π$ against the first strategy of player II yields a payoff $$π^TMe_1=p^T0+q^T\left(-A^T_{\cdot 1}\right)+0\cdot c_1=-q^TA_{\cdot 1}$$ where with $A^T_{\cdot 1}$ we denote the first column of the matrix $A^T$. Now this payoff is negative since all the elements of $-A$ are negative unless $q$ is the zero vector. But if $q$ is the zero vector then the payoff of $π$ against the last strategy of Player II is equal to $$π^ΤΜe_{m+n+1}=-p^Tb+0<0$$ which is a contradiction. So $r>0$. Similarly we can show more, i.e. that $p,q$ are not the zero vectors.
For 3. Use that $π$ guarantees to Player I his security level against any strategy of Player II, i.e. that $π^ΤΜe_j\ge 0$ for every $j=1,2,\ldots, m+n+1$. This gives the set of inequalities $$\begin{cases}-q^TA^T+rb^T\ge0\\p^TA-rc^T\ge0\\-p^Tb+q^Tc\ge0\end{cases} \implies \begin{cases}\dfrac{q^T}{r}A^T\le b^T\\\dfrac{p^T}{r}A\ge c^T\\\dfrac{p^T}{r}b\le \dfrac{q^T}{r}c\end{cases}$$ The first inequality gives you that the vector $\dfrac{q^T}{r}$ is primal feasible, the second inequality that the vector $\dfrac{p^T}{r}$ is dual feasible and the third that the value of the objective function of the primal LP for the primal feasible vector $\dfrac{q^T}{r}$ is greater or equal than the value of the objective function of the dual LP for the dual feasible vector $\dfrac{p^T}{r}$. Due to strong duality (actually weak duality suffices) this yields optimality of the solutions.