This is about problem 4.12 from Bertsimas & Tsitsiklis's Introduction to Linear Optimization à la page 190. The problem is repeated below (bold is from errata).
Consider a general linear programming problem and suppose that we have a non-degenerate basic feasible solution to the primal in which all free variables are nonzero. Show that the complementary slackness conditions lead to a system of equations for the dual vector that has a unique solution.
From my understanding, a general LP primal-dual pair is as follows.
\begin{equation*} \begin{array}{cc} \begin{array}{rll} \text{minimize} & \textbf{c}^\top\textbf{x}&\\ \text{subject to} & \textbf{a}_i^\top\textbf{x}\geqslant b_i,&i\in M_1\\ & \textbf{a}_i^\top\textbf{x}\leqslant b_i,&i\in M_2\\ & \textbf{a}_i^\top\textbf{x}= b_i,&i\in M_3\\ & x_j\geqslant 0,&j\in N_1\\ & x_j\leqslant 0,&j\in N_2\\ & x_j~\text{free},&j\in N_3\\ \end{array} & \begin{array}{rll} \text{maximize} & \textbf{p}^\top\textbf{b}\\ \text{subject to} & p_i\geqslant 0,&i\in M_1\\ & p_i\leqslant 0,&i\in M_2\\ & p_i~\text{free},&i\in M_3\\ & \textbf{p}^\top\textbf{A}_j\leqslant c_j,&j\in N_1\\ & \textbf{p}^\top\textbf{A}_j\geqslant c_j,&j\in N_2\\ & \textbf{p}^\top\textbf{A}_j= c_j,&j\in N_3\\ \end{array} \\ \end{array} \end{equation*}
A basic feasible solution $\mathbf{x}^*$ would imply that at such a point, there are exactly $n$ linearly independent active constraints. This does not say that $|M_3|=n$. There could be non-free variables that are $0$. However, this UCBerkeley solution claims so and the professor marked it correct. Could anyone explain why $|M_3|=n$ and why there cannot be any $x_i=0$ in this case?
The solution that you have linked is wrong. Notice that the book you are using explicitly gives a definition of what a general optimization problem looks like on page 42: \begin{equation*} \begin{array}{ll} \text{minimize} \quad & \mathbf{c}^{\prime}\mathbf{x}\\ \text{subject to} \quad & \mathbf{A}\mathbf{x} \geq \mathbf{b} \end{array} \end{equation*}
To answer your second question, we have $x_i \neq 0$ for all indices of $x$ (and not only for some subset of $\{1,\ldots,n\}$).
You can find a correct solution to this exercise here.