Computing cohomology over projective curve in $\mathbf{P}^3$ Let be $k$ an algebraically closed field and Let be $X\subseteq \mathbf{P}^3:=\mathbf{P}_k^3$ a smooth, irreducible curve that is not contained in any hyperplane. Let's call $d=\deg(X)$.
A well known theorem of Gruson, Lazarsfeld and Peskine states that such a curve is $d-(3-1)+1=d-1$-regular, that is 
$$H^p(\mathbf{P}^3,\mathscr{I}_X(d-1-p))=0$$
for all $p>0$. Now, some simple computations on this definition show us that this reduces to the following conditions:
$$H^1(\mathbf{P}^3,\mathscr{I}_X(d-2))=0,\,\,\,H^1(\mathbf{P}^3,\mathscr{O}_X(d-3))=0$$
Now the actual question. I would like to show that the theorem holds without using it, but I'm a bit struck. I need in particular the cases of a complete intersection curve and a rational curve of degree $3$.
For example, if you write $X=F_1\cap F_2$ as complete intersection of hypersurfaces $F_1,F_2$ of degree $d_1,d_2$ respectively, then the Koszul complex (aka Hilbert-Burch complex) resolves the ideal of $X$:
$$0\to \mathscr O_{\mathbf{P}^3}(-d_1-d_2)\to \mathscr O_{\mathbf{R}^3}(-d_1)\oplus \mathscr O_{\mathbf{R}^3}(-d_1)\to \mathscr{I}_X\to 0$$
Switching to cohomology sequence, this should imply $H^1(\mathbf{P}^3,\mathscr{I}_X(s))=0$ for all $s$; I'm not so sure this is completely correct.
As for the other condition, we may compute the canonical bundle $$\omega_X=\omega_{\mathbb{P}^3}(+d_1+d_2)=\omega_{\mathbb{P}^3}(d_1+d_2-4)$$ so when $s\geq d_1+d_2-3$ we have $H^1(\mathscr{O}_X(s))=0$ for sure. Then the inequality $d=d_1d_2\geq d_1+d_2-1$ shows the second condition. 
Can somebody help me understand better these computations? How should I proceed to yeld a similar result in the case of a rational curve of degree $3$?
 A: Your computation in the complete intersection case are correct, I just want to point out that you can work just with the Koszul complex, using the first definition of regularity.
When $X$ is the rational normal curve of degree $3$ we can proceed like this: we can look at $X$ as the embedding of $\mathbb{P}^1$ induced by the complete linear system $\left|\mathcal{O}_{\mathbb{P}^1}(3)\right|$, and then we have the identifications $H^i(\mathbb{P}^3,\mathcal{O}_{X}(d)) = H^i(\mathbb{P}^1,\mathcal{O}_{\mathbb{P}^1}(3d))$. In particular $$H^1(\mathbb{P}^3,\mathcal{O}_X(d-3)) = H^1(\mathbb{P}^3,\mathcal{O}_X) = H^1(\mathbb{P}^1,\mathcal{O}_{\mathbb{P}^1})=0$$.
For the other condition, consider the exact sequence of sheaves
$$ 0 \to \mathscr{I}_X(1) \to \mathcal{O}_{\mathbb{P}^3}(1) \to \mathcal{O}_X(1) \to 0 $$
taking cohomology we get the exact sequence
$$ H^0(\mathbb{P}^3,\mathcal{O}_{\mathbb{P}^3}(1)) \to H^0(\mathbb{P}^1,\mathcal{O}_{\mathbb{P}^1}(3)) \to H^1(\mathscr{I}_X(1)) \to 0 $$
however, the map $H^0(\mathbb{P}^3,\mathcal{O}_{\mathbb{P}^3}(1)) \to H^0(\mathbb{P}^1,\mathcal{O}_{\mathbb{P}^1}(3))$ is an isomorphism, since $X$ is the embedding of $\mathbb{P}^1$ by the complete linear system $\left| \mathcal{O}_{\mathbb{P}^1}(3) \right|$, so that $H^1(\mathscr{I}_X(1))=0$  
