Q2: Which pair to remove from $R$ to make it an equivalence relation

An equivalence relation must be reflexive, symmetric and transitive, so we test each property in turn.

Reflexive on S means $(a, a) \in R$ for every $a$ in $S = \{1, 2, 3\}$⁠. All three such pairs are in R.

Symmetric means: whenever $(a, b) \in R$⁠, also $(b, a) \in R$⁠.

The only pair with two different entries is $(1, 2)$⁠, and its reverse $(2, 1)$ is missing. That one pair is the problem, so it is the element to remove.

Every pair left has the form $(a, a)$⁠. Its reverse is itself, so R is symmetric. Joining $(a, a)$ with $(a, a)$ gives $(a, a)$ again, so R is transitive.

Watch out: removing a pair such as $(1, 1)$ instead would break reflexivity, and R would still not be symmetric. So $(1, 2)$ is the only element to remove.