Let $Y$ be a smooth cubic 4-fold in $\mathbf{P}^5$. The derived category of $Y$ admits a semiorthogonal decomposition

$$D^b(Y) = \langle \mathcal{A}_Y, \mathcal{O}_Y, \mathcal{O}_{Y}(1), \mathcal{O}_{Y}(2) \rangle,$$

where $\mathcal{A}_Y$ is the right-orthogonal to the triangulated subcategory generated by $\mathcal{O}_Y, \mathcal{O}_{Y}(1), \mathcal{O}_{Y}(2)$. Namely, $\mathcal{A}_Y$ is the full subcategory with objects those $F \in D^b(Y)$ such that $Hom_{D^b(Y)}(\mathcal{O}_Y(j), F[i]) = 0$ for $j=0,1,2$ and all $i$.

Kuznetsov studies $\mathcal{A}_Y$ in his paper "Derived categories of cubic fourfolds" http://arxiv.org/abs/0808.3351. He parenthetically remarks that the Serre functor of $\mathcal A_Y$ is the shift by two functor $[2]$.

The question is: How do you show this?

The only reference I could find for this result is Corollary 4.4 of the paper http://arxiv.org/abs/math/0303037. However, I don't understand the proof given there. Specifically, I don't know how to justify the following points in Lemma 4.2 preceding the Corollary (notation as in the paper):

1) How do you get the expression Kuznetsov gives for the kernel of $O^d$ from the kernel of $O$?

2) Why is the restriction to $Y \times Y$ of the Beilinson resolution of the diagonal in $\mathbf{P}^{n+1} \times \mathbf{P}^{n+1}$ quasi-isomorphic to ${\Delta_{Y}}_*(\mathcal O_{Y})$? It seems the paper is being sloppy here with what degrees the complexes live in: Let's call $A$ the restriction of the Beilinson resolution (including the final term ${\Delta_{Y}}_*(\mathcal O_{Y}(d))$ as in the paper). Let's place A in nonpositive degrees so that ${\Delta_{Y}}_*(\mathcal O_{Y}(d))$ is in degree $0$ (I assume this is what is intended in the paper). Then I think the precise statement should be that $A$ is quasi-isomorphic to ${\Delta_{Y}}_*(\mathcal O_{Y})[2]$. Moreover, since $A$ is the restriction of a resolution of $\Delta_*(\mathcal O_{\mathbf{P}^{n+1}}(d))$ by locally free sheaves, this can be rephrased as the following computation of left-derived functors: if $i : Y \times Y \hookrightarrow \mathbf{P}^{n+1} \times \mathbf{P}^{n+1}$ is the inclusion, then $$L_1 i^*(\Delta_*(\mathcal O_{\mathbf{P}^{n+1}}(d))) = {\Delta_{Y}}_*(\mathcal O_{Y}) $$ and $$L_k i^*(\Delta_*(\mathcal O_{\mathbf{P}^{n+1}}(d))) = 0$$ for $k > 1$. How do you show this?