Here is the exercise and its solution:
1-Is there a typo and $\varphi$ should be $\psi$?
2- I do not understand how by exercise 2.1.14 we will get the first equation in the solution of 4.1.32. And how this is precisely the statement that the diagram below it commutes?
Could someone explain the above questions to me please?
$$ \psi_{A,B} \colon \mathcal{B}(F(A), B) \to \mathcal{A}(A, G(B)), $$ and we consider $f \colon A \to G(B)$, i.e., $f \in \mathcal{A}(A, G(B))$. It is possible that $\varphi_{A,B}$ is supposed to denote the inverse of $\psi_{A,B}$, because it is applied to $f$. But in the diagram, I think it should still be $\psi_{A,B}$. This could also be a reason why everything that follows is not so clear.
Let me assume that $f \in \mathcal{B}(F(A), B)$ (contrary to what is done in the solution). Then, $\psi_{A,B}(f) \in \mathcal{A}(A, G(B))$. This is really just a change in notation since $\psi_{A,B}$ is a bijection.
$$ \psi_{A,B} \colon \mathcal{B}(F(A), B) \to \mathcal{A}(A, G(B)). $$
If we have $f \in \mathcal{B}(F(A), B)$, then we can denote the image of $f$ under this map $\psi_{A,B}$ as we usually do by $\psi_{A,B}(f)$. This is how it is written in the statement of Exercise 4.1.32.
But we can also adopt the notation $\bar{f}$ for the image of $f$ under $\psi_{A,B}$ and then
$$ \bar{f} = \psi_{A,B}(f). $$
(see Definition 2.1.1). This notation appears in Exercise 2.1.14. (This may be a bit confusing since one needs to watch out which way the arrows go. Sketching a diagram might be a good idea at this point.)
If we translate the conclusion of Exercise 2.1.14. into the notation used in 4.1.32., the equality reads
$$ \psi_{A', B'}(q \circ f \circ F(p)) = G(q) \circ \psi_{A,B}(f) \circ p. $$
So we indeed recover the first statement of the solution.
It remains to see why this equality is exactly the statement that the given diagram commutes. For this, one needs to remember the definition of $\mathcal{A}(p, G(q))$ (and analogously $\mathcal{B}(F(p), q)$). This can be seen in Remark 4.1.24. and Definition 4.1.22. The map is in fact quite simple. We have $$ \mathcal{A}(p, G(q)) \colon \mathcal{A}(A,G(B)) \to \mathcal{A}(A', G(B')), \\ \alpha \mapsto G(q) \circ \alpha \circ p. $$ In words, we compose the map $\alpha$ from the left with $G(q)$ and from the right with $p$.
So say we start at the top left of the diagram with any $f \in \mathcal{B}(F(A), B)$, then we can chase this element through the diagram and see that the diagram commutes if and only if we have
$$ \psi_{A', B'}(q \circ f \circ F(p)) = G(q) \circ \psi_{A,B}(f) \circ p. $$
Since this diagram is a naturality square, this then shows that the isomorphism is natural.
I hope all my arrows point in the right direction. Let me know if you find an issue somewhere.