Arrange according to cardinality: $\ \Bbb{R}\setminus\Bbb{N}, P(\{1\}), P(\Bbb{Q}), \Bbb{R}\setminus\{π,e\}, P(\{2,5\}), \{3,7\}, \Bbb{R}\setminus\Bbb{Z} , \Bbb{R}\setminus\Bbb{Q}$(irrationals)
So far I believe I know cardinalities : $\ \{3,7\} < P(\{1\}) <\cdots $
An important observation is that the cardinality of the power set of a set with cardinality $n \in \mathbb N \ \text{is} \ 2^n.$ Hence, the cardinality of $ P(\{1\})$ is $2^1=2$, and is equal to the cardinality of $ \{3,7\}$. By the same money: $ \vert P({2,5})\vert = 4$. (Here $ \vert S\vert$ refers to the cardinality of the set S).
Now, $\mathbb R \setminus \mathbb N$ is uncountable, and so are P($\mathbb Q$), $ \ $ $\mathbb R \setminus \{π,e\}$ and $\mathbb R \setminus \mathbb Z$. Yet another important theorem is Cantor's power set theorem which states that for any gives set S, |S|$\lt$|P(S)| - i.e. the cardinality of the power set of S is strictly greater than that of S. Hence, by the above theorem we can conclude that |P($\mathbb R \setminus \mathbb Q$|$\gt$ |$\mathbb R \setminus \mathbb Q$| = |$\mathbb R$|.
Finally, putting everything together:
$$|P(\mathbb R \setminus \mathbb Q)| \gt |\mathbb R \setminus \mathbb Z| \ = \ |\mathbb R \setminus \mathbb N| = |\mathbb R \setminus \{π,e \}| \ = \ |P(\mathbb Q)| \gt |P(\{2,5 \})| \ = \ 4 \gt |P(\{1 \})| = 2 = |\{3,7 \}|.$$