How I can find all equivalence classes by Myhill-Nerode?
$\Bbb N$ denotes the set of natural numbers, that is, the set $\{0,1,2, \cdots\}$. If $\Bbb N$ is the set of positive integers, the treatment will be roughly the same.




  So in $1$ they build a class of the empty word $\lambda$ and $z = B$ has to be the language by her self to be in the language?


Since the above paragraph in the question, $z=z(C)$ is understood/redefined to be the set of words such that Cz is a subset of $B$, given a set of words $C$. That is, $cw\in b$ for any $c\in C$ and $w\in z$.

With the above understanding, yes, $z([\lambda]_{\equiv_{B}})=z(\{\lambda\})=B.$


  $3: \left[73 a^{k}\right]_{\equiv_{B}}=\left\{73 a^{k}\right\}  $ for $  k \in \mathbb{N} $


Case 3 gives one equivalent class for each $k\in\Bbb N$. It lists infinitely many equivalent classes as the following, each with different $z$.


Letting $k=0$, we have $\left[73\right]_{\equiv_{B}}=\left\{73\right\}\,.$
Letting $k=1$, we have $\left[73 a^1\right]_{\equiv_{B}}=\left\{73 a^1\right\}\,.$
Letting $k=2$, we have $\left[73 a^2\right]_{\equiv_{B}}=\left\{73 a^2\right\}\,.$
Letting $k=3$, we have $\left[73 a^3\right]_{\equiv_{B}}=\left\{73 a^3\right\}\,.$
$\vdots$


Their $z$'s are the following, which are more or less listed in the question.


$z\left(\left[73\right]_{\equiv_{B}}\right)=z\left(\left\{73 \right\}\right)=\left\{a^n7b^m \mid n,m\in \Bbb{N} \wedge n=m+2\right\} \,.$
$z\left(\left[73 a^1\right]_{\equiv_{B}}\right)=z\left(\left\{73 a^1\right\}\right)= \left\{a^n7b^m \mid n,m\in \Bbb{N} \wedge n=m+1\right\} \,.$
$z\left(\left[73 a^2\right]_{\equiv_{B}}\right)=z\left(\left\{73 a^2\right\}\right)=\left\{a^n7b^m \mid n,m\in \Bbb{N} \wedge n=m\right\} \,.$
$z\left(\left[73 a^3\right]_{\equiv_{B}}\right)=z\left(\left\{73 a^3\right\}\right)=\left\{a^n7b^m \mid n,m\in \Bbb{N} \wedge n=m-1\right\} \,.$
$\vdots$


Once you have understood the above, it should be easy to figure out case 4 and the rest of the given solution. You did have a pretty good understanding of the  Myhill-Nerode equivalence classes.



Here is another way to write the same solution in the question.

$1: [\lambda]_{\equiv_{B}}=\{\lambda\}  $

$2: [7]_{\equiv_{B}}=\{7\}  $

$3: \left[73\right]_{\equiv_{B}}=\left\{73\right\}$

$4: \left[73a\right]_{\equiv_{B}}=\left\{73a\right\}$

$5: \left[73a^2\right]_{\equiv_{B}}=\left\{73a^2\right\}$

$6: \left[73a^3\right]_{\equiv_{B}}=\left\{73a^3\right\}$

$\vdots$

$\infty+1: \left[73 a^2 7\right]_{\equiv_{B}} =\left\{73 a^{2+n} 7 b^{n} \mid n \in \mathbb{N}\right\}$ 

$\infty+2: \left[73 a^3 7\right]_{\equiv_{B}} =\left\{73 a^{3+n} 7 b^{n} \mid n \in \mathbb{N}\right\}$ 

$\infty+3: \left[73 a^4 7\right]_{\equiv_{B}} =\left\{73 a^{4+n} 7 b^{n} \mid n \in \mathbb{N}\right\}$ 

$\infty+4: \left[73 a^5 7\right]_{\equiv_{B}} =\left\{73 a^{5+n} 7 b^{n} \mid n \in \mathbb{N}\right\}$ 

$\vdots$

$2\infty+1: [3]_{{\equiv B}}=\Sigma^{*} \setminus\left([\lambda]_{\equiv B} \cup[7]_{\equiv B}  \cup\left(\bigcup_{k \in \Bbb{N}}\left[73 a^{k}\right]_ {\equiv_{B}}\right) \cup \left(\bigcup_{k \in \Bbb{N}}\left[73a^{k+ 2} 7\right]_{\equiv_B}\right)\right)\,. $



Exercise. Given the language 
$$ A \triangleq\left\{1^{n} 0^{n} \mid n \in \mathbb{N}\right\} \text { with } \Sigma_{A} \triangleq\{1,0\}, $$
give all equivalence classes of the Myhill-Nerode relation. (Check here for an answer.)