(i) \(\{x\in N; x^2 < 25\} = \{1,2,3,4\}\)

\(\because 1^2 = 1 < 25, 2^2 = 4<25, 3^2 = 9< 25, 4^2 = 16 < 25 \;\text{but}\; 5^2 = 25\)

(ii) \(\{x\in N; x\) is a prime number and \(10 < x < 20\}\)

\(\because \) 11, 13, 17 & 19 are prime numbers between 10 & 20.

\(\{x\in N; x\) is a prime number and \(10 < x < 20\} = \{11, 13, 17, 19\}\) 

(iii) 17, 26, 35, 44, 53, 62 & 71 are two digit numbers whose sum of digits is 8.

\(\therefore\) {x : x is a two digit number such that the sum of its digits is 8} = {17, 26, 35, 44, 53, 62, 71}.

(iv) 9, 18, 27, 36, 45, ...... are positive integer which are divisor of 9.

\(\therefore\) {x : x is a positive integer and a divisor of 9} = {9, 18, 27, 36, 45, .....}

(v) \(|x| \le 2\)

⇒ \(-2 \le x \le 2\)

-2, -1, 0, 1 & 2 are integer which belongs to \(-2 \le x \le 2\).

\(\therefore \{x: x\in Z\;and\;|x| \le 2\}= \{-2, -1, 0, 1, 2\}\)

(ix) \(a_{n +2} = a_{n + 1} + a_n \) & \(a_1 = a_2 = 1\)

\(\therefore a_3 = a_2 + a_1= 1+1=2\)

\(a_4 = a_3+a_2 = 2+1= 3\)

\(a_5 = a_4+ a_3 = 3+2 = 50\)

\(a_6 = a_5 + a_4 = 5 + 3= 8\)

\(\therefore \{a_n : n\in N, a_{n + 2} = a_{n + 1} + a_n \;and\; a_1 = a_2 = 1\}= \{1,1,2,3,5,8,...\}\)

(x) \(a_{n + 1} = 3a_n \) & \(a_1 = 2\)

\(\therefore a_2 = 3a_1 = 3 \times 2 = 6 = 2\times 3\)

\(a_3 = 3a_2 = 3 \times 6 = 18 = 2\times 3^2\)

\(a_4 = 3a_3 = 3\times 18 = 54= 2 \times 3^3\)

\(a_n = 2\times 3^{n -1}\)

\(\therefore \{a_n : n\in N, a_{n + 1} = 3a_n \; and\; a_1 = 2\}= \{{2, 6, 18, 54,..., 2\times 3^{n -1},...\}}\)