( i) A={x|x is prime factor of 65 } = {2,3}
B={ x| x is solution os `x^2- 5x + 6 = 0 }`
`x^2- 5 x + 6 =0`
or ( x-2)( x-3)=0
or x=2, 3
`therefore ` B= { 2,3}
Therefore , A+B
(ii) A={x|x is a letter in the word REPLACED
= { R,E,P,L,A,C,E,D}
B= { y |y is a letter in the word PARCELED}
B= { P,A,R,C,E,L,E,D}
Therefore , A= B.
(iii) A= { x|x is natural number , `x gt 1} = { 2, 3,4, ..... }`
B+ { x|x is natural number `x ge 1} = { 1,2,3 ...3}`
We observe that `1 in B "but " 1 notin A`
Therefore , `A ne B`