$aRb = { (a,b): 5a\equiv 2b (mod3) }$
1. Set containing all elements containing at least 1 3s as multiple.
From relation, $5a - 2b \equiv 0(mod 3)$
$\Rightarrow 5a - 2b = 3k$ for some k.
Since a & b are multiple of 3. For all a & b We can have,
$\Rightarrow 3(5a' - 2b') = 3k$
Which means for every a & b. Given relation is being satisfied therefore a & b in the class.
2. Counter example S = { 2,4 }
3. Counter example S = { 1,3}
4. Counter example S = { 2,4}