\(X\cap Y=\{ b,d\} \)

(A\(\cap \)U) = {2, 4, 7} \(\cap \) {2, 3, 4, 5, 6, 7, 8, 9, 10, 11}
={2, 4, 7}
B\(\cup \)C ={3, 5, 7, 9, 11} \(\cup \) {7, 8, 9, 10, 11}
(A\(\cap \)U)\(\cap \)(B\(\cup \)C) = {2, 4, 7} \(\cap \) {3, 5, 7, 8, 9, 10, 11}
= {7}

{3, 6, 15, 18, 21}

{0,1,2}

(A \(\cap\) B) \(\cup\) (A - B)
= (A \(\cap\) B) \(\cup\) (A \(\cap\) B') = A \(\cap\) (B \(\cup\) B') (By distributive law)
=A\(\cap\)U=A
Hence A = (A\(\cap\) B) \(\cup\) (A - B)
Also A \(\cup\) (B - A)
=A\(\cup\)(B\(\cap\)A')
= (A \(\cup\) B) \(\cap\) (A \(\cup\) A') (By distributive law)
= (A\(\cup\)B) \(\cap\) U
=A\(\cup\)B
Hence A \(\cup\) (B - A) = A \(\cup\) B.

(i) A' = U-A = {a, b, c, d, e, f, g, h} - (a, b, c}
= {d, e, f, g, h}
(ii) B' = U - B = {a, b, c, d, e, f, g, h} - {d, e, f, g}
= {a, b, c, h}
(iii) C' = U - C = {a, b, c, d, e, f, g, h} - {a, c, e, g} = {b, d, f, h}
(iv) D' = U - D = {a, b, c, d, e, f, g, h} - {f, g, h, a}
= {b, c, d, e}

Here
A = {x : x is a natural number} = {1, 2, 3, 4, 5,......}
B = {x : x is an even natural number} = {2, 4, 6,.......}
C = {x : x is a an odd natural number} = {1, 3, 5, 7,......}
and D= {x : x is a prime number} = {2, 3, 5, 7,.....}
(i) A \(\cap\) B = {x : x is a natural number} \(\cap\){x:x is an evennatural number}
= {x: x is an even natural number}
=B
(ii)A\(\cap\) C = {x : x is a natural number} \(\cap\)(x: x is an oddnatural number}
= {x: x is an odd natural number}
=C
(iii) A\(\cap\)D ={x : x is a natural number} \(\cap\) (x : x is a prime number}
= {x : x is a prime number}
=D
(iv) B\(\cap\)C = {x: xis an even natural number} \(\cap\){x: xis an oddnatural number}
=ф
(v) B \(\cap\)D = {x: xis an even natural number} \(\cap\){x : x is a prime number}
= {2}
(vi) C \(\cap\) D ={x :x is an odd natural number} \(\cap\) {x : x is a prime number}
= {x: x is an odd prime number}

(i) {3, 4} is a member of set A.
\(\therefore\) {3, 4} ∈ A
Hence {3, 4}⊂ A is incorrect.
(ii) {3, 4} is a member of set A.
\(\therefore\){3, 4} ∈ A is correct.
(iii) Here {3, 4} is a member of set A.
\(\therefore\) {{3,4}} is a set
\(\therefore\) {{3,4}} ⊂ A is correct.
(iv) 1 is a member of set A.
\(\therefore\) 1 ∈ A is correct.
(v) 1 is not a set, it is a member of set A.
\(\therefore\) 1⊂ A is incorrect.
(vi) 1, 2, 5 are members of set A.
\(\therefore\) {1, 2, 5} is a subset of set A.
\(\therefore\) {1, 2, 5} ⊂ A is correct.
(vii) 1, 2, 5 are members of set A.
\(\therefore\) {1, 2, 5} is a subset of set A.
\(\therefore\) {1, 2, 5} ∈ A is incorrect.
(viii) 3 is not a member of set A.
\(\therefore\) {1, 2, 3} is not a subset of set A.
\(\therefore\) {1, 2, 3} ⊂ A is incorrect.
(ix) ф is not a member of set A.
\(\therefore\) ф ∈ A is incorrect.
(x) Since ф is subset of every set,
\(\therefore\)ф ⊂ A is correct.
(xi)ф​​​​​​​ is not a member of set A.
\(\therefore\) {ф} ⊂ A is incorrect.