This is tricky - the two cases look the same, but they're not. The first one is a right action $v \cdot (p_1 \cdot p_2) = (v \cdot p_1) \cdot p_2$, while the second one is a left action $p_1 \cdot (p_2 \cdot f) = (p_1 \cdot p_2) \cdot f$. To see why, consider these 2 permutations:
$$p_1(1) = 1, p_1(2) = 3, p_1(3) = 2 $$
and
$$p_2(1) = 3, p_2(2) = 2, p_2(3) = 1.$$
Let's write out explicitly what the actions are in the two cases to see the difference.
First, let's work out what the compositions $p_1 \cdot p_2$ and $p_2 \cdot p_1$ are.
The composition $p_1 \cdot p_2$ is:
$$p_1(p_2(1)) = 2, p_1(p_2(2)) = 3, p_1(p_2(3)) = 1$$
while the composition $p_2 \cdot p_1$ is:
$$p_2(p_1(1)) = 3, p_2(p_1(2)) = 1, p_2(p_1(3)) = 2.$$
Observe that they are not the same. We will use these later.
Now let's look at the 2 actions. The first action is on vectors. By the definition of the first action, $$p_1 \cdot (v_1, v_2, v_3) = (v_{p_1(1)}, v_{p_1(2)}, v_{p_1(3)}) = (v_1, v_3, v_2).$$ In words: $p_1$ acting on a vector interchanges the second and third coordinates. Similarly, $$p_2 \cdot (v_1, v_2, v_3) = (v_3, v_2, v_1)$$ In words: $p_2$ acting on a vector interchanges the first and third coordinates.
(In this situation, I find that thinking in words reduces the confusion: $p_1$ interchanges the second and third coordinates, not $v_2$ and $v_3$. You'll see the difference below.)
Thus, $$p_1 \cdot (p_2 \cdot v) = p_1 \cdot (p_2 \cdot (v_1, v_2, v_3)) = p_1 \cdot (v_3, v_2, v_1) = (v_3, v_1, v_2).$$ (If the last equality seems wrong, use the "words" description of $p_1$: Interchange the second and third coordinates.) Now the rightmost term above is $(p_2 \cdot p_1) \cdot v$, not $(p_1 \cdot p_2) \cdot v$. (Use the calculation of $p_2 \cdot p_1$ above.)
Conclusion: For the first action, on vectors, $p_1 \cdot (p_2 \cdot v) = (p_2 \cdot p_1) \cdot v$. So this is not a left action, but a right action.
Now look at the second action, which is not on vectors, but on real-valued functions on the set of all vectors. By definition, $$(p_1 \cdot f)(x_1, x_2, x_3) = f(x_{p_1(1)}, x_{p_1(2)}, x_{p_1(3)}) = f(x_1, x_3, x_2).$$
In words: To evaluate $p_1$ applied to a function at a vector, interchange the second and third coordinates of the vector, then apply the function.
Similarly, for $p_2$, the words version is: To evaluate $p_2$ applied to a function at a vector, interchange the first and third coordinates of the vector, then apply the function.
So what is $p_1 \cdot (p_2 \cdot f)$, applied to a vector? It is $$(p_1 \cdot (p_2 \cdot f))(x_1, x_2, x_3) = (p_2 \cdot f)(x_1, x_3, x_2) = f(x_2, x_3, x_1).$$
(Again, using the "words" description may reduce the confusion.) Now is this last term $p_1 \cdot p_2$ or $p_2 \cdot p_1$ applied to $f$? It is the former, as you can see from the calculation of $p_1 \cdot p_2$ above.
Conclusion: $p_1 \cdot (p_2 \cdot f) = (p_1 \cdot p_2) \cdot f$. This one is a left action.
I hope this clears up why these two cases aren't the same. Of course, I haven't yet proven that these are actions in general (I've only illustrated it for the particular $p_1$ and $p_2$ above), but hopefully this will give you the right idea for the general proof.
The key point is to remember that with these definitions, we are permuting the coordinates of the vector according to $p$, rather than the indices of the $v$'s according to $p$. If we did the latter instead, then the left and right action cases above would be reversed. See alias vs alibi for more discussion of this point.
It really doesn’t make sense to talk about the values of $x$ for which the action is faithful: faithfulness is a property of the action as a whole. I can’t make any sense of $$\implies g_2^{-1}g_1\;:$$ $g_2^{-1}g_1$ isn’t something that can be implies. It’s an element of the group $G$, not a statement.
Suppose that $\bigcap_{g\in G}gHg^{-1}=\{1_G\}$. In order to show that the action is faithful, you must show that if $g_1,g_2\in G$, and $g_1gH=g_2gH$ for all $g\in G$, then $g_1=g_2$. Now
$$\begin{align*}
g_1gH=g_2gH\quad&\text{ iff }\quad g^{-1}g_2^{-1}g_1g\in H\\
&\text{ iff }\quad g_2^{-1}g_1\in gHg^{-1}\;,
\end{align*}\tag{1}$$
so if $g_1gH=g_2gH$ for all $g\in G$, we must have
$$g_2^{-1}g_1\in\bigcap_{g\in G}gHg^{-1}\;.$$
If $\bigcap_{g\in G}gHg^{-1}=\{1_G\}$, this implies that $g_2^{-1}g_1=1_G$ and hence that $g_1=g_2$, as desired.
Conversely, suppose that $\bigcap_{g\in G}gHg^{-1}\ne\{1_G\}$, and fix $g_1\in\bigcap_{g\in G}gHg^{-1}$ with $g_1\ne 1_G$. Let $g_2=1_G$; you can reverse the calculations above to show that $g_1gH=g_2gH$ for all $g\in G$ and hence that the action is not faithful: $g_1$ and $1_G$ act identically on $G/H$.
Best Answer
For $1$ and $2$ take $g=ax^{-1}$ for whichever $a\in G$ you want.
For $3$, $\{g : x^{-1}gx\in H \} = \{g : gx\in xH \} =\{g : g\in xHx^{-1} \}=xHx^{-1}$.
For $4$, if $H\not= G$, we can find an $a\notin H$ and look at the coset $aH$. Then $a^{-1}$ sends $aH$ to $H$, and similarly $a$ sends $H$ to $aH$, so no coset can be fixed under the action of $G$.
Another alternative way to see $4$ would be to use $3$. Take any coset $xH$, and we have that $\text{Stab}_G(xH)=xHx^{-1}$. Since $xHx^{-1}$ has order $|H|$, if $|H|<|G|$, then $$|\mathcal{O}_{xH}|=[G:\text{Stab}_G(xH)]=[G:H]>1$$ so $xH$ is not a fixed point.