Consider $X$ smooth projective curve such that $X\subseteq \mathbb P^3$, $X$ is not contained in any hyperplane, and $g(X)=1$. In this exercise the idea is to consider the long exact sequence we get from $$0\to \mathcal I_X(2)\to \mathcal O_{\mathbb P^3}(2)\to \mathcal O_X(2)\to 0,$$ and then say that $\dim H^0(\mathbb P^3,\mathcal O_{\mathbb P^3}(2))=10$ and $\dim H^0(X,\mathcal O_X(2))=8$, and we deduce that $\dim H^0(X,\mathcal I_X(2))\geq 2$.
My question is why do we have $\dim H^0(X,\mathcal O_X(2))=8$?
Let $C$ be a smooth projective curve (curve includes irreducible here), and let $D$ be a very ample divisor of degree $d$. Then the global sections of $\mathcal{O}_C(D)$ give a closed immersion $i_D:C\to\Bbb P^N$ which expresses $i_D(C)$ as a curve of degree $d$: by construction of the embedding, $i_D^*\mathcal{O}_{\Bbb P^N}(1)\cong \mathcal{O}_C(D)$.
All of this is reversible: if $C$ has a closed immersion in to $\Bbb P^N$ by some map $i:C\to\Bbb P^N$, then $i^*\mathcal{O}_{\Bbb P^N}(1)$ is a very ample sheaf on $C$ by definition, and it must be the line bundle associated to a divisor (since $C$ is integral, every line bundle is the line bundle associated to a divisor). Which divisor? Well, any hyperplane $H\subset\Bbb P^n$ must intersect $C$, and if it does not contain $C$ it gives us a collection of $\deg_{\Bbb P^N} C$ points, counted with multiplicity, and this divisor works! This is what we mean when we say "the hyperplane divisor" on a curve embedded in $\Bbb P^N$ (just to be extra clear, this is only a divisor up to linear equivalence), and it's what we're looking at when we take something like $\mathcal{O}_C(2)$: we're really taking $i^*\mathcal{O}_{\Bbb P^N}(2)$, so we'll get $\mathcal{O}_C(2(H\cap C))$.
In your specific case, you have a curve of genus 1 and degree 4. Then the degree of the hyperplane divisor is the degree of the curve, or 4, so the degree of twice the hyperplane divisor is 8. By Riemann-Roch, on a curve of genus 1, a divisor of degree 8 has an 8-dimensional space of global sections.