Solution:

1. We will solve it by two ways; the first way is the usual axioms check. Clearly, $\mathcal{S} \neq \emptyset $. Let $x^2, y^2 \in \mathcal{S} $, we need to show that $x^2 \cdot y^2 \in \mathcal{S} $. Since $G$ is abelian, we have \[ x^2 \cdot y^2 = (x\cdot y)^2 \in \mathcal{S} . \] Now we need to show that it is closed under inversion, that is if $x\in \mathcal{S} $, then $x^{-1} \in \mathcal{S} $. Let $x^2 \in \mathcal{S} $. Note that $x\in G$ implies $x^{-1} \in G$. So \[ \left( x^{-1} \right)^2 \cdot x^2 = e = x^2 \left( x^{-1} \right) ^2. \] Therefore, $\left( x^{-1} \right) ^2 $ is the inverse of $x^2$ and it belongs to the set $\mathcal{S} $ as $x^{-1} \in G$. Therefore, $\mathcal{S} $ is a subgroup of $G$.
   
   We can also show the same by considering the map \[ \varphi : G \to G,~ x \mapsto x^2. \] Then \[ \varphi \left( xy \right) = (xy)^2 = x^2 y^2 = \varphi (x) \varphi (y) \] and hence, $\varphi $ is a group homomorphism. Note that \[ \mathcal{S} = \operatorname{im}\left( \varphi \right) \] and hence $\mathcal{S} $ is a subgroup of $G$.
2. We need to show that the index of $\mathcal{S} $ in $G$ is $2$. Note that from the first isomorphism theorem we have \[ G/\ker \varphi \cong \operatorname{im}\left( \varphi \right) = \mathcal{S} . \] Note that if $x\in \ker \varphi $, then $x^2 \equiv 1 $ mod $p$, which implies $(x-1)(x+1) \equiv 1 $ mod $p$. Since $\mathbb{Z} /p\mathbb{Z} $ is an integral domain, $x \equiv \pm 1$ mod $p$. Therefore, $\ker \varphi = \{ -1,1 \}$. So, \[ \left\vert \mathcal{S} \right\vert = \left\vert G \right\vert / \left\vert \ker \varphi \right\vert = \frac{p-1}{2}. \] Hence, \[ [G:\mathcal{S} ] = \frac{\left\vert G \right\vert }{\left\vert \mathcal{S} \right\vert } = 2. \]
3. Since $-1 \notin \mathcal{S} $ and $[G:\mathcal{S} ] = 2$, so we can write \[ G = \mathcal{S} \sqcup \left( -\mathcal{S} \right), \] where $\sqcup$ denotes the disjoint union. If $a\in G\setminus \mathcal{S} $. Since $a \notin \mathcal{S} $ but in $G$ so $a \in -\mathcal{S} $. Therefore, there exists $b\in \mathcal{S} $ such that $a = -b$ which implies $b = -a \in \mathcal{S} $. Thus we proved that exactly one of $a$ or $-a$ will be in $\mathcal{S}$.