$${R_1}$$ is not reflexive, because $$\left( {q,\,q} \right)\left( {r,\,r} \right)\, \notin \,{R_1}$$
$$\therefore \,{R_1}$$  is not equivalence relation
$${R_2}$$ is not reflexive, because $$\left( {p,\,p} \right)\, \notin \,{R_2}$$
$$\therefore \,{R_2}$$  is not equivalence relation
$${R_3}$$ is reflexive, because $$\left( {p,\,p} \right),\,\left( {q,\,q} \right),\,\left( {r,\,r} \right)\, \in \,{R_3}$$
$${R_3}$$ is not symmetric, because $$\left( {p,\,q} \right)\, \in \,{R_3}$$   but $$\left( {q,\,p} \right)\, \notin \,{R_3}.$$

$$\eqalign{ & A \cup B = A \cr & \Rightarrow B \subseteq A \cr & \Rightarrow A \cap B = B \cr} $$

$$x R y$$  need not implies $$y R x$$
∴ $$R$$ is not symmetric and hence not an equivalence relation.
$$\eqalign{ & S:\frac{m}{n}S\frac{p}{q} \cr & {\text{Given}}\,{\text{ }}qm = pn \cr & \Rightarrow \,\,\frac{p}{q} = \frac{m}{n} \cr & \therefore \,\,\frac{m}{n}S\frac{m}{n}\left( {{\text{reflexive}}} \right)\frac{m}{n}S\frac{p}{q} \cr & \Rightarrow \,\,\frac{p}{q}S\frac{m}{n}\left( {{\text{symmetric}}} \right)\frac{m}{n}S\frac{p}{q},\frac{p}{q}S\frac{r}{s} \cr & \Rightarrow \,qm = pn,ps = rq \cr & \Rightarrow \,\,\frac{p}{q} = \frac{m}{n} = \frac{r}{s} \cr & \Rightarrow \,\,ms = rn\left( {{\text{transitive}}} \right). \cr} $$
$$S$$ is an equivalence relation.

$$\eqalign{ & n\left( A \right) = 2{\text{ and }}n\left( B \right) = 2 \cr & n\left( {A \times B} \right) = n\left( A \right) \times n\left( B \right) = 2 \times 2 = 4 \cr & \therefore \,{\text{Number of subsets of }}A \times B = {2^{n\left( {A \times B} \right)}} = {2^4} = 16 \cr} $$

Let $$U$$ denote the set of surveyed students and $$X$$ denotes the set of students taking apple juice and $$Y$$ denote the set of students taking orange juice. Then, $$n\left( U \right) = 400,\,\,n\left( X \right) = 100,\,\,n\left( Y \right) = 150$$        and $$n\left( {X \cap Y} \right) = 75$$
$$\eqalign{ & n\left( {X \cup Y} \right) = n\left( X \right) + n\left( Y \right) - n\left( {X \cap Y} \right) \cr & = 100 + 150 - 75 \cr & = 175 \cr} $$
$$\therefore \,175$$   students were taking at least one juice.
$$\eqalign{ & n\left( {X' \cap Y'} \right) = n\left( {X \cup Y} \right)' \cr & = n\left( U \right) - n\left( {X \cup Y} \right) \cr & = 400 - 175 \cr & = 225 \cr} $$
Hence, 225 students were taking neither apple juice nor orange juice.

$$\eqalign{ & {\text{We have,}} \cr & R = \left\{ {\left( {a,\,b} \right):a + b\, \in \,Z,\,a,\,b\, \in \,Z \cup \left\{ {\sqrt 2 } \right\}} \right\} \cr & \sqrt 2 \, \in \,A\,{\text{ and }}\sqrt 2 + \sqrt 2 = 2\sqrt 2 \, \notin \,Z \cr & \therefore \,\left( {\sqrt 2 ,\,\sqrt 2 } \right) \notin \,Z\,\,\,\,\,\therefore \,R{\text{ is not reflexive}}{\text{.}} \cr & {\text{Let }}\left( {a,\,b} \right)\, \in \,R\,\,\,\,\,\therefore \,a + b\, \in \,Z \cr & \Rightarrow b + a\, \in \,Z \cr & \Rightarrow \left( {b,\,a} \right) \in \,R\,\,\,\,\, \Rightarrow R{\text{ is symmetric}}{\text{.}} \cr & {\text{Let }}\left( {a,\,b} \right),\,\left( {b,\,c} \right)\, \in \,R \cr & \therefore \,a + b,\,b + c\, \in \,Z \cr & \therefore \,{\text{None of }}a,\,b,\,c\,{\text{is equal to }}\sqrt 2 \cr & \therefore \,a,\,b,\,c\, \in \,Z \cr & \therefore \,a + c\, \in \,Z \cr & \Rightarrow \left( {a,\,c} \right) \in \,R\,\,\,\,\, \Rightarrow R{\text{ is transitive}}{\text{.}} \cr & \therefore \,R{\text{ is not an equivalence relation}}{\text{.}} \cr} $$

Given
$$n\left( A \right) = 2,n\left( B \right) = 4,n\left( {A \times B} \right) = 8$$
Required number of subsets $$ = {\,^8}{C_3} + {\,^8}{C_4} + ..... + {\,^8}{C_8}$$
$$\eqalign{ & = {2^8} - {\,^8}{C_0} - {\,^8}{C_1} - {\,^8}{C_2} \cr & = 256 - 1 - 8 - 28 \cr & = 219 \cr} $$

$$\eqalign{ & \left( {2,\,3} \right)\, \in \,R{\text{ but }}\left( {3,\,2} \right)\, \notin \,R \cr & \therefore \,R{\text{ is not symmentric}}{\text{.}} \cr} $$

$$\eqalign{ & {\text{Since }}\mu \,{\text{is universal set and }}P \subseteq \mu ,\,P - \mu = \phi \,\,{\text{and }}\mu - P = P' \cr & {\text{So, }}\left( {P - \mu } \right) \cup \left( {\mu - P} \right) = \phi \cup P' = P' \cr & {\text{Now, }}P \cap \left\{ {\left( {P - \mu } \right) \cup \left( {\mu - P} \right)} \right\} = P \cap P' = \phi \cr} $$

If $$A$$ is a subset of the universal set $$U,$$ then its complement $$A'$$ is also a subset of $$U.$$
We have, $$A' = \left\{ {2,\,4,\,6,\,8,\,10} \right\}$$
Hence, $$\left( {A'} \right)' = \left\{ {x:x\, \in \,U{\text{ and }}x\, \notin \,A'} \right\} = \left\{ {1,\,3,\,5,\,7,\,9} \right\} = A$$
It is clear from the definition of the complement that for any subset of the universal set $$U,$$ we have $$\left( {A'} \right)' = A$$