Algebraic property of a set

Line by line...

$A$ and $B$ are subsets of $U=\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ such that

${ A }^{ c }\cap { B }^{ { c } }=\{1, 9\}$ --- means that neither 9 nor 1 is in A or B, and everything except 9 and 1 are in at least one of A or B.

$A\cap B =\{2\}$ --- means that 2 is the only thing in both A and B, leaving 3, 4, 5, 6, 7, and 8 to be in one of A or B but not both.

${ A }^{ c }\cap B=\{4, 6, 8\},$ --- means that 4, 6, and 8 are the things in B but not A, leaving 3, 5, and 7 to be in A but not B.

So, in total, 2, 3, 5, and 7 are in A.

Solution by answer choices:

You only need one of the three clues. When it's learned 4, 6, and 8 are all part of the complement of A in the third clue, three answer choices are ruled out.

$(A^c\cap B)\cup (A^c\cap B^c) = A^c\cap (B\cup B^c) = A^c\cap U = A^c \implies A^c = \{ 1,9\}\cup \{4,6,8\} \implies A= \{2,3,5,7\}$