1) Let $A = \{0, 1, 2, 3, 4 \}$. Define the relation $M$ from $P(A)$ to $A$ in the following way: $(X, x) \in M$ if and only if $x = \min X$.
2) Define the relation $L \subseteq \mathbb{N} \times \mathbb{R}$ in the following way: $x\mathrel{L}y$ if and only if $x = y^3$.
3) Define the relation $Q$ on $\mathbb{Q}$ in the following way: $(a/b, c/d) \in Q$ if and only if $ad = bc$, where we assume $a$, $c \in \mathbb{Z}$, and $b$, $d \in \mathbb{N}$.
I'm very lost as for what the answers should be. For 2 I tried domain as all real numbers and range as natural number but that didn't make much sense to me. As for 1 and 3 I'm fully lost.
Let $R \subseteq A \times B$ be a relation:
1) $M = \{(X, x) \in \mathcal{P}(A) \times A : x = \min X \}$. Every subset of $A$ except $\emptyset$ has a minimum, hence the domain is $\mathcal{P}(A)\setminus \emptyset$ and its range is $A$, for every element $a \in A$ is the minimum of $\{a\} \in \mathcal{P}(A)$.
2) $L$ is the set of ordered pairs $(x,y)$ of $\mathbb{N} \times \mathbb{R}$ that satisfy the equation $x = y^3$. Let $(x,y) \in L$, $x$ can be any natural number because there always exists $y \in \mathbb{R}$ such that $y^3 = x$ ($x \in \mathbb{N} \Rightarrow (\sqrt[3]{x})^3 = x$). This tells us the domain is $\mathbb{N}$. Its range is not $\mathbb{R}$, for example, there is no $x \in \mathbb{N}$ such that $x = \pi^3$. The range is the set $\{ \sqrt[3]{n} : n \in \mathbb{N} \} \subset \mathbb{R}$.
3) This is the way equality is defined when building up $\mathbb{Q}$. $\frac{a}{b} = \frac{c}{d}$ if and only if $ad = bc$. Every fraction is equal to some other fraction, ($\frac{a}{b} = \frac{q \cdot c}{q \cdot d}$ for every $q \in \mathbb{Q}$), which implies the domain is $\mathbb{Q}$. Since this is an equivalence relationship its range is also $\mathbb{Q}$.