Let us remember the definition of faithful, transitive and free:
The action of a group $G$ on a set $X$ is called:
Transitive: if $X$ is non-empty and if for any $x,y\in X$ there exists $g\in G$ such that $g\cdot x = y$.
Faithful (or effective): if for any two distinct $g,h\in G$ there exists $x\in X$ such that $g\cdot x \neq h\cdot x$; or equivalently, if for any $g\neq e\in G$ there exists $x\in X$ such that $g\cdot x \neq x$. Intuitively, in a faithful group action, different elements of $G$ induce different permutations of $X$.
Free: if, given $g,h\in G$, the existence of $x\in X$ with $g\cdot x = h\cdot x$ implies $g = h$. Equivalently: if $g$ is a group element and there exists an $x\in X$ with $g\cdot x = x$ (that is, if $g$ has at least one fixed point), then $g$ is the identity.
Now we can determine your three group actions:
The action of $S_n$ on a set of $n$-elements is as you showed transitive. It is also faithful as you explained, but it is not free, because there are elements on $S_n$ that do have fixed points that are not the identity, for example $1\to 1, 2\to 3, 3\to 4, n-1\to n, n\to 2$. Since the action is transitive the group of orbits is the group of one element.
The action of $O(n)$ on $S^{n-1}$, it is transitive, since there is always a rotation that can send any point of $S^{n-1}$ onto the north pole. For $n>2$, It is not free since for any point on $S^{n-1}$ there are many rotations different from the identity, whose axis of rotation is the line that passes through that point and the origin, and all of these fix that point. It is not faithful neither since any element of $O(n)$ that is not the identity moves $S^{n-1}$, therefore there exists a point on $S^{n-1}$ that is not fixed by the action. Since the action is transitive the group of orbits is the group of one element. The case $n=2$ the action it is transitive, free but not faithful.
The action of $O(n)$ on $\mathbb{R}^n$ is not free since every element of $O(n)$ fixes the origin but is not necessarily the identity. It is not transitive either, since for two elements of $\mathbb{R}^n$ whose length are not equal, you cannot find an element of $O(n)$ that sends one on the other (the elements of $O(n)$ preserve norm). It is not faithful neither since when you restrict the action to $S^{n-1}$ it is not faithful either. Since $\mathbb{R}^n=[0,\infty)\times S^{n-1}$ the set of orbits can be identified with $[0,\infty)$ since when $O(n)$ is restricted to the second factor it is transitive.
I think the link you provided proves the last part correctly. The key is recognizing the correspondence between block and subgroups. Fixing $a \in A$, then there is a correspondence between
\begin{align*}
\{\text{subgroups } H \text{ with } G_a \subseteq H \subseteq G \} & \longleftrightarrow
\{\text{subsets } B \text{ with } \{a\} \subseteq B \subseteq A\}\\
H &\longmapsto \{\sigma(a) : \sigma \in H\}\\
G_B = \{\sigma \in G : \sigma(B) = B\} \ &{\longleftarrow\!\shortmid} \ B
\end{align*}
which is proven in your link. We can prove part (d) using this fact.
($\Leftarrow$): To complete your argument, note that since the blocks partition $A$, we can choose a block $B$ that contains $a$. You already covered the case $G_B = G$, so suppose $G_B = G_a$. For contradiction, assume $B \neq \{a\}$ so that there is some $b \neq a$ with $b \in B$. Since $G$ is transitive, then there exists $\sigma \in G$ such that $\sigma(a) = b$. Then $b \in \sigma(B) \cap B$, so $\sigma(B) = B$. Then $\sigma \in G_B$ but $\sigma \notin G_a$, contradiction.
($\Rightarrow$): Suppose that $G$ is primitive and there is a subgroup $H$ with $G_a \subseteq H \subseteq G$. Let $B = \{\sigma(a) : \sigma \in H\}$. Since $G$ is primitive, then either $B = \{a\}$ or $B = G$. In the first case ($B = \{a\}$), then every element of $H$ stabilizes $a$, so $H = G_a$. In the latter case ($B = G$), then $H = G_B = G_A = G$ since the link proved $H = G_B$.
Best Answer
(0). Exercise. Explain how equivalence relations, partitions, and homomorphic images of a set are all equivalent concepts. If you can understand why they're pretty much the same for sets, you'll be in a better position to understand the generalization to $G$-sets.
(1) If $G$ acts on a set $\Omega$, then it acts on subsets too. For instance, for any $k$, it acts on the collection of $k$-subsets of $\Omega$, (by defining $gA:=\{ga\mid a\in A\}$ for subsets $A\subseteq\Omega$). It also acts on the collection of set-theoretic partitions of $\Omega$. For instance, let $S_3$ act on $\Omega=\{1,2,3,4\}$ in the obvious way. Then, picking a permutation $g=(123)$ and partition $\pi=\{\{1,2\},\{3,4\}\}$ of $\Omega$, we have
$$ g\pi=\{\{2,3\}.\{1,4\}\}. $$
A $G$-stable partition of $\Omega$ is a fixed point of this action. For instance if $H=V_4$, comprised of the identity and permutatons of cycle type $(ab)(cd)$, then $\pi$ is $H$-stable, but e.g. $\{\{1,2\},\{3\},\{4\}\}$ isn't.
(2) Suppose $\Omega$ and $\Psi$ are $G$-sets. A function $\phi:\Omega\to\Psi$ is called $G$-equivariant if it intertwines two actions, i.e. $\phi(g\omega)=g\phi(\omega)$ for all $g\in G$ and $\omega\in\Omega$. It is also called a $G$-morphism.
In this way, there is a category of $G$-sets just like there are categories of sets (functions), groups, rings (homomorphisms), vector spaces (linear transformations), topological spaces (continuous functions), etc.
"Onto" means surjective, i.e. for all $\psi\in\Psi$ there exists at least one $\omega\in\Omega$ for which $\phi(\omega)=\psi$.
(3) For two $G$-sets $\Omega$ and $\Psi$, if there exists an onto $G$-morphism $\phi:\Omega\to\Psi$, we call $\Psi$ a homomorphic image of $\Omega$. That is, $\Psi$ is the image (aka range) of a $G$-morphism.
"Nontrivial" means not the trivial $G$-set, i.e. $|\Psi|>1$. (The one-element trivial $G$-set is trivially a homomorphic image of any other $G$-set.) I suppose we should also exclude $\Omega$; every $G$-set is trivial image of itself (with $\phi$ the identity map).
There is the same concept for plain old groups - for two groups $G$ and $H$, if there is an onto group homomorphism $G\to H$ we call $H$ a homomorphic image of $G$.
(4). Exercise. A group is simple if and only if it has no nontrivial homomorphic images.
(5). Fibers of any $G$-morphism are blocks.
(6). Primitivity is useful in verifying many of the classified simple groups are indeed simple.
Blocks can also be useful "transversals" for orbits that help us visualize how a group acts. The first three pictures of Starting with the group $SL_2(\mathbb{R})$ come to mind.