According to 'Distribution law' in set theory the given both statements are wrong.
Statement 1 - Distribution law
Statement 2 - Distribution law

Since, Power set is the collection of all the subsets of the set $$A$$ therefore $$AUP\left( A \right) = P\left( A \right)$$
$$\therefore $$  statement (1) is correct.

Since, $$A$$ is void set therefore the number of elements in power set of $$A$$ is $$1.$$
$$\therefore \,P\left\{ {P\left( A \right)} \right\} = {2^1} = 2$$
because, if set $$A$$ has $$n$$ elements then $$P\left( A \right)$$  has $${2^n}$$ elements.
$$\eqalign{ & \Rightarrow \,P\left\{ {P\left\{ {P\left( A \right)} \right\}} \right\} = {2^2} = 4 \cr & \Rightarrow P\left\{ {P\left\{ {P\left\{ {P\left( A \right)} \right\}} \right\}} \right\}\, = {2^4} = 16 \cr} $$

$$\eqalign{ & \left( {\text{A}} \right)\,\,A = \left\{ {x:x\, \in \,Z{\text{ and }}{x^2} - 5x + 6 = 0} \right\} = \left\{ {2,\,3} \right\} \cr & {\text{So, }}A{\text{ is a finite set}}{\text{.}} \cr & \left( {\text{B}} \right)\,\,B = \left\{ {x:x\, \in \,Z{\text{ and }}{x^2}{\text{ is even}}} \right\} \cr & = \left\{ {.....,\, - 6,\, - 4,\, - 2,\,0,\,2,\,4,\,6,.....} \right\} \cr & {\text{Clearly, }}B\,{\text{is an infinite set}}{\text{.}} \cr & \left( {\text{C}} \right)\,\,D = \left\{ {x:x\, \in \,Z{\text{ and }}x > - 10} \right\} \cr & = \left\{ { - 9,\, - 8,\, - 7,.....} \right\} \cr & {\text{Clearly, }}D\,{\text{is an infinite set}}{\text{.}} \cr} $$

Here $$R = \left\{ {\left( {x,\,y} \right):\left| {{x^2} - {y^2}} \right| < 16} \right\}$$      and given $$A = \left\{ {1,\,2,\,3,\,4,\,5} \right\}$$
$$\therefore \,R = \left\{ {\left( {1,\,2} \right)\left( {1,\,3} \right)\left( {1,\,4} \right);\,\left( {2,\,1} \right)\left( {2,\,2} \right)\left( {2,\,3} \right)\left( {2,\,4} \right);\left( {3,\,1} \right);\left( {3,\,2} \right)\left( {3,\,3} \right)\left( {3,\,4} \right);\left( {4,\,1} \right)\left( {4,\,2} \right)\left( {4,\,3} \right);\left( {4,\,4} \right),\left( {4,\,5} \right),\left( {5,\,4} \right)\left( {5,\,5} \right)} \right\}$$

$${\left( {A - B} \right) \cup \left( {B - C} \right) \cup \left( {C - A} \right)}$$      is represented by the shaded portion in the figure. The unshaded portion is $$A \cap B \cap C.$$
$$\therefore \left\{ {\left( {A - B} \right) \cup \left( {B - C} \right) \cup \left( {C - A} \right)} \right\}' = A \cap B \cap C.$$

$$\eqalign{ & R = \left\{ {\left( {1,\,1} \right),\,\left( {1,\,2} \right),\,\left( {2,\,1} \right),\,\left( {2,\,2} \right),\,\left( {2,\,3} \right),\,\left( {3,\,2} \right),\,\left( {3,\,3} \right)} \right\} \cr & {R^{ - 1}} = \left\{ {\left( {y,\,x} \right):\left( {x,\,y} \right)\, \in \,R} \right\} \cr & = \left\{ {\left( {1,\,1} \right),\,\left( {2,\,1} \right),\,\left( {1,\,2} \right),\,\left( {2,\,2} \right),\,\left( {3,\,2} \right),\,\left( {2,\,3} \right),\,\left( {3,\,3} \right)} \right\} \cr & = R \cr & {\text{Domain of }}R = \left\{ {x:\left( {x,\,y} \right)\, \in \,R} \right\} = \left\{ {1,\,2,\,3} \right\} \cr & {\text{Range of }}R = \left\{ {y:\left( {x,\,y} \right)\, \in \,R} \right\}\left\{ {1,\,2,\,3} \right\} \cr} $$

$$\eqalign{ & {R_1}{\text{ is a relation from }}X{\text{ to }}Y{\text{ because }}{R_1} \subseteq X \times Y \cr & {R_2}{\text{ is a relation from }}X{\text{ to }}Y{\text{ because }}{R_2} \subseteq X \times Y \cr & {R_3}{\text{ is a relation from }}X{\text{ to }}Y{\text{ because }}{R_3} \subseteq X \times Y \cr & {R_4}{\text{ is not a relation from }}X{\text{ to }}Y{\text{ because }}\left( {2,\,4} \right),\left( {7,\,9} \right) \notin \,X \times Y \cr} $$

If either $$P$$ or $$Q$$ is the null set, then $$P \times Q$$  will be an empty set, i.e., $$P \times Q = \phi .$$

$$\eqalign{ & {\text{We have,}} \cr & gof\left( 2 \right) = g\left( {f\left( 2 \right)} \right) = g\left( 3 \right) = 7, \cr & gof\left( 3 \right) = g\left( {f\left( 3 \right)} \right) = g\left( 4 \right) = 7, \cr & gof\left( 4 \right) = g\left( {f\left( 4 \right)} \right) = g\left( 5 \right) = 11, \cr & {\text{and }}gof\left( 5 \right) = g\left( 5 \right) = 11 \cr} $$