Consider the collection of $m$ pairwise disjoint lines $L_1,\ldots,L_m$ in $\mathbb{P}^3$ and pose $Z=L_1\sqcup\cdots\sqcup L_m$. Consider the rank-$2$ vector bundle on $\mathbb{P}^3$ which is given by the extension $$0\to\mathcal{O}_{\mathbb{P}^3}\to E\to I_Z(2)\to0 \tag 1$$ where $I_Z$ is the ideal sheaf of $Z$. I want to compute $H^i(\mathbb{P}^3, E(k))$ for all $i$ and $k\in\mathbb{Z}$.
Here is my try, which is probably not the best possible. Twisting (1) by $k$ we obtain the exact sequence $$0\to\mathcal{O}_{\mathbb{P}^3}(k)\to E(k)\to I_Z(k+2)\to0.\tag 2$$ Then, taking the cohomologies of (2) we obtain that
1) For $k\geq0$, $h^0(E(k))=h^0(\mathcal{O}_{\mathbb{P}^3}(k))+h^0(I_Z(k+2))=\binom{k+3}3+h^0(I_Z(k+2))$ and $h^i(E(k))=h^i(I_Z(k+2))$, $i=1,2,3$.
2) For $k<0$, $h^i(E(k))=h^i(I_Z(k+2))$, $i=0,1$ and we have the exact sequence $$0\to H^2(\mathbb{P}^3, E(k))\to H^2(\mathbb{P}^3, I_Z(k+2))\to H^3(\mathbb{P}^3, \mathcal{O}_{\mathbb{P}^3}(k))\to H^3(\mathbb{P}^3, E(k))\to H^3(\mathbb{P}^3, I_Z(k+2))\to0.$$
Thus everything is boiled down to computing $H^i(\mathbb{P}^3, I_Z(n))$ for $n\in\mathbb{Z}$. Let us compute this for $n\geq0$, which is the hardest part actually (for $n<0$ this is easy).
We have the exact sequence $$0\to I_Z(n)\to\mathcal{O}_{\mathbb{P}^3}(n)\to\mathcal{O}_{Z}(n)\to0.$$ Taking the cohomologies of it we obtain that $h^i(I_Z(n))=0$ for $i=2,3$ and the the exact sequence $$0\to H^0(\mathbb{P}^3, I_Z(n))\to H^0(\mathbb{P}^3, \mathcal{O}_{\mathbb{P}^3}(n))\stackrel{\alpha}\to H^0(\mathbb{P}^3, \mathcal{O}_Z(n))\to H^1(\mathbb{P}^3, I_Z(n))\to0.\tag 3$$
Note that since $L_1,\ldots,L_m$ are pairwise disjoint, $\mathcal{O}_Z=\bigoplus\limits_{i=1}^m\mathcal{O}_{L_i}$. For $n=0$ the map $\alpha$ in (3) is clearly an injection. Also it is easy to see that for $m=1,2$ and all $n$ the map $\alpha$ is a surjection, so it is possible to compute $h^i(I_Z(n))$ in these cases.
For $m>2$ it doesn't seem possible to say something about this map. Could anyone help me?