My book is Connections, Curvature, and Characteristic Classes by Loring W. Tu (I'll call this Volume 3), a sequel to both Differential Forms in Algebraic Topology by Loring W. Tu and Raoul Bott (Volume 2) and An Introduction to Manifolds by Loring W. Tu (Volume 1).
I refer to Section 27.1.
Let $M$ be a set, possibly empty. Let $G$ be a group, possibly a singleton. Let $G$ act right on $M$ by the action $\mu: M \times G \to M$. For each $x \in M$, let $\text{Stab}(x):=\{g \in G | \mu(x,g) = x\}$ denote a stabilizer subgroup of $G$. Let $1_G$ be the identity of $G$.
I understand definitions of $\mu$ to be free as follows:
Wikipedia: $\mu$ is free if for all $g \in G$, if there exists $x \in M$ such that $\mu(x,g)=x$, then we have that $g=1_G$.
jgon in this answer: (same as Wikipedia's, given above)
Section 27.1: $\mu$ is free if for all $x \in M$, $\text{Stab}(x) = \{1_G\}$
Question 1: For Wikipedia's and jgon's definitions, there is no explicit reference to stabilizers. For Tu's definition, how do I understand $\text{Stab}(x)$ for $M$ empty and $G$ not a singleton?
Question 2: Similarly, for the definition of faithful as
$$\bigcap_{x \in M} \text{Stab}(x) = \{1_G\} \tag{2a}$$
How do I understand $\mu$ as never faithful for $M$ empty and $G$ not a singleton?
My attempt to understand:
For Question 2, I think I can apply this, by $M$ empty assumption to say $\bigcap_{x \in M} \text{Stab}(x) = G$. Then I apply $G$ not a singleton assumption to get $\bigcap_{x \in M} \text{Stab}(x) \ne \{1_G\}$.
For Question 1, I think we somehow say $\text{Stab}(x) = G$ for all $x \in M = \emptyset$ by some vacuousness argument. I'm not really sure.
$G$ acts on $M$ faithfully iff the induced homomorphism $\psi$ from $G$ to the symmetric group on $M$ is injective. If $M$ is empty then the kernel of this homomorphism is all of $G$ (because the symmetric group on the empty set is trivial), so the action is faithful iff $G$ is trivial.
$G$ acts on $M$ freely iff every stabilizer is trivial. If $M$ is empty, there are no stabilizers to speak of, so the action is free.