Let $X:=\mathbb{P}^r_k$ and $Y$ a closed subscheme of dimension $q$ which is a complete intersection. The reader is asked to show $H^i(Y,\mathcal{O}_Y(n))=0$ for all $0<i<q$ and all $n\in\mathbb{Z}$.
The following fake proof yields too much information, and my suspicion is that the initial step is incorrect. I would like an explanation on what is wrong. I have bolded the part that I suspect is faulty. Indeed, I think the cohomology of $i_*\mathcal{O}_Y(n)$ may not be the cohomology group $H^i(Y,\mathcal{O}_Y(n))$...
Look at the short exact sequence $0\rightarrow \mathscr{I}_Y\rightarrow \mathcal{O}_X\rightarrow i_*\mathcal{O}_Y\rightarrow 0$ where $i:Y\hookrightarrow X$. Now twist this by $n$ to get $0\rightarrow \mathscr{I}_Y(n)\rightarrow \mathcal{O}_X(n)\rightarrow i_*\mathcal{O}_Y(n)\rightarrow 0$. Take the long exact sequence of cohomology,
$$0\rightarrow H^0(X,\mathscr{I}_Y(n))\rightarrow H^0(X,\mathcal{O}_X)\rightarrow\dots H^i(X,\mathscr{I}_Y(n))\rightarrow H^i(X,\mathcal{O}_X(n))\rightarrow H^i(Y,\mathcal{O}_Y(n))\rightarrow H^{i+1}(X,\mathscr{I}_Y(n))\rightarrow\dots $$
Now the sheaf $\mathscr{I}_Y(n)$ is quasi-coherent since $\mathscr{I}_Y$ is quasi-coherent and $\mathcal{O}_X(n)$ is invertible. So Serre's Criterion says all higher cohomology vanishes. Since we know $H^i(X,\mathcal{O}_X(n))=0$ for all $0<i<r$, we then obtain $H^i(Y,\mathcal{O}_Y(n))=0$ for all $0<i<r$ as well.
Edit: This method would also give part (a) of the problem for free, but again, it appears faulty.
This is a very short answer which I realized from a comment of @Karl Kroningfeld
The solution is incorrect because Serre's Criterion requires that scheme to be affine. This is not true in this case.