(Most of this was written before the recent addendum. It addresses the OP's original question, not the addendum.)
(a) Suppose we have distinct bases $B_1$ and $B_2$ that each yield the same basic solution ${\bf x}$. Now, suppose (we're looking for a contradiction) that ${\bf x}$ is nondegenerate; i.e., every one of the $m$ variables in ${\bf x}$ is nonzero. Thus every one of the $m$ variables in $B_1$ is nonzero, and every one of the $m$ variables in $B_2$ is nonzero. Since $B_1$ and $B_2$ are distinct, there is at least one variable in $B_1$ not in $B_2$. But this yields at least $m+1$ nonzero variables in ${\bf x}$, which is a contradiction. Thus ${\bf x}$ must be degenerate.
(b) No. The counterexample linked to by the OP involves the system $$
\begin{align}
x_1 + x_2 + x_3 = 1, \\
-x_1 + x_2 + x_3 = 1, \\
x_1, x_2, x_3 \geq 0.
\end{align}$$
There are three potential bases in this system: $B_1 = \{x_1, x_2\}$, $B_2 = \{x_1, x_3\}$, $B_3 = \{x_2, x_3\}$. However, $B_3$ can't actually be a basis because the corresponding matrix $\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$ isn't invertible. $B_1$ yields the basic solution $(0,1,0)$, and $B_2$ yields the basic solution $(0,0,1)$. Both of these are degenerate, but there is only one basis corresponding to each.
(c) No. Look at the system
$$
\begin{align}
x_1 + x_2 = 1, \\
x_2 + x_3 = 1, \\
x_1, x_2, x_3 \geq 0.
\end{align}
$$
The basic solution $(0,1,0)$ corresponds to bases $\{x_1, x_2\}$ and $\{x_2, x_3\}$. The only other basis is $\{x_1, x_3\}$, which implies that the only other basic solution is $(1,0,1)$. Thus the degenerate basic solution $(0,1,0)$ is not adjacent to another degenerate basic solution.
(
More on part (a), addressing OP's questions in the comments.)
Say there are $n$ total variables in the problem: $x_1, x_2, \ldots, x_n$. Every basis $B$ consists of some $m$ of these variables. The basic solution ${\bf x}$ corresponding to a given basis $B$ has the other $n-m$ variables equal to $0$. (Setting these to $0$ is partly how you determine the value of ${\bf x}$; see, for instance, the examples above). If ${\bf x}$ is degenerate it might have some of the variables in $B$ equal to $0$, too, but the point in terms of the argument is that ${\bf x}$ can have no more than $m$ nonzero variables.
Now, suppose $B_1$ and $B_2$ are distinct and each have $m$ nonzero variables, yet both correspond to ${\bf x}$. Let's say $B_2 = \{x_1, x_2, \ldots, x_m\}$. Since $B_1$ and $B_2$ are distinct, $B_1$ has at least one variable that's not in $B_2$. Let's say this variable is $x_{m+1}$. But since every variable in $B_1$ and $B_2$ is nonzero, that means that $x_1, x_2, \ldots, x_m, x_{m+1}$ are all nonzero. However, $B_1$ and $B_2$ both correspond to ${\bf x}$, which means that there are at least $m+1$ nonzero variables in ${\bf x}$. That cannot happen for a basic solution, and so we have a contradiction.
Indeed, this has to do with the fact that we are pivoting from one extreme point to another. When you are on an extreme point, necessarily some of your variables equal $0$, do you see why ?
In two dimensions, you need at least two lines to define a point. In terms of linear programming, an extreme point is at the intersection between two constraints that are active. And for a constraint to be active, necessarily the corresponding slack variable equals $0$. So in two dimensions, you need at least two variables to have value $0$ to be on an extreme point (and the other variables to be non negative).
Best Answer
Yes to both questions. I first start with a simple example, then ellaborate the definition of degeneracy.
A simple degenerate LPP
To illustrate this problem, let me use this example.
$\max x$ subject to
\begin{align} \color{blue}{x+y}&\le\color{blue}{1}\\ \color{red}x\phantom{+y}&\le\color{red}1\\ x,y&\ge0 \end{align}
Obviously, exactly one of the blue and red constraints is redundant, so this LPP has degenerate solution.
We transform it to the standard form by adding slack variable $\color{blue}{s_1}, \color{red}{s_2}$.
$\max x$ subject to
\begin{align} \color{blue}{x+y+s_1\phantom{+s_2}}&=\color{blue}{1}\\ \color{red}{x\phantom{+y+s_1}+s_2}&=\color{red}1\\ x,y,\color{blue}{s_1},\color{red}{s_2}&\ge0 \end{align}
Thus, each of $x=0,y=0,\color{blue}{s_1=0},\color{red}{s_2=0}$ corresponds to a line which bounds the feasible region.
Therefore, in the first prompt, the condition (zero-valued slack variable $s_i$ in the basic variable $x_B$) is irrelevant to the degeneracy of $x_B$. The definition of degeneracy still applies to $x_B=(1,\color{blue}0)$ and $x_B=(1,\color{red}0)$.
Theoretical stuff
Suppose $\mathrm{rank}(B)=n$. You have at most $n-1$ variables with non-zero value in the current solution. The degeneracy of a solution just depends on the presence of zero entry in the basic solution $B^{-1}b$. It doesn't matter whether the basic variable is a slack variable or not.