Sheaves on $\mathbb{P}^n \times \mathbb{P}^m$, and a commutation relation for derived functors of global sections and tensor products on it. I'll state my questions first and then provide some background. Question 3 is by far my most important one. We work over $k=\mathbb{C}$ whenever necessary.


*

*Is it true that $\text{Pic}(\mathbb{P}^n \times \mathbb{P}^m) \cong \mathbb{Z} \oplus \mathbb{Z}$, with generators $\pi_1^*(\mathcal{O}_{\mathbb{P}^n}(1))$ and $\pi_2^*(\mathcal{O}_{\mathbb{P}^m}(1))$? (here the $\pi_i$ are the canonical projections).

*Is it true that the presheaf $U \mapsto  \pi_1^*(\mathcal{O}_{\mathbb{P}^n}(k_1))(U) \otimes  \pi_2^*(\mathcal{O}_{\mathbb{P}^m}(k_2))(U)$ is already a sheaf for all $k_1, k_2 \in \mathbb{Z}$? If not, is this true if we demand the $k_i$ to be positive, or equal, or both?

*Let $F, G$ be line bundles on a projective variety $X$ such that their tensor product in presheaves is already a sheaf. Then of course $H^0(X, F \otimes G) \cong H^0(X,F) \otimes H^0(X,G)$, differently stated the global sections functor commutes with the tensor product (although the tensor products are "different", one is is (pre)sheaves and one is in $k$-vector spaces).
Question: Is there a similar relation for the higher sheaf cohomology functors commuting with the tensor product? I can imagine that there exists some spectral sequence for this, most probably also involving the higher Tor functors.
Background:
I am considering a singular surface in $\mathbb{P}^3$ and its desingularization in the blowup $S \subset \tilde{\mathbb{P}^3} \subset \mathbb{P}^3 \times \mathbb{P}^2$. I am considering some particular line bundle $Q$ on $S$ and am intensely interested in its first cohomology group. I have little information to work with, but after tensoring this sheaf with some other "easy" bundles its cohomology can be easily calculated. I am hoping i can translate this back to information about the bundle $Q$.
My attempts so far:
For 1., i am pretty sure Hartshorne II.6.6 and II.6.6.1 generalize directly but it feels slippery, which is why i ask. About 2, when thinking of 
$$
\mathbb{P}^n \times \mathbb{P}^m = \text{Proj}(k[z_{ij}]_{i,j})/\sim
$$
and Serre's construction of the twisting sheaves where the global sections are just elements of this ring of some degree, it seems to be true at least for global sections and $k_1 = k_2$. Again, it feels slippery enough to ask.
As for 3 i have no idea (yet). 
Please feel free to use the language of schemes and homological algebra, i am supposed to be familiar with this.. Thanks a lot!
 A: Partial answer. 
1) Yes. Maybe the simplest way is to view an element of the Picard group as an equivalence class of Weil divisors. Fix an affine space $U_n$ (resp. $V_m$) in $\mathbb P^n$ (resp. $\mathbb P^m$) with complement $H_n, L_m$ isomorphic to a projective space of dimension one lower. Let $D$ be a Weil divisor on $\mathbb P^n\times \mathbb P^m$. Then $D|_{U_n\times V_m} \sim 0$ because $U_n\times V_m$ is defined by a UFD domain. This implies that up to linear equivalence, $D$ is supported in the complement of $U_n\times V_m$ which is $E_1:=H_n\times \mathbb P^m$ union $E_2:=V_m\times \mathbb P^n$. That is $D=k_1E_1+k_2E_2$ with $k_i\in\mathbb Z$. In terms of elements in Picard, this means $O(D)\simeq \pi_1^*(O_{\mathbb P^n}(k_1))\otimes \pi_2^*(O_{\mathbb P^m}(k_2)).$
2) True if $U$ is affine (trivial) or if $U$ is a product of open subsets (baby step of Künneth's formula, easily proved by direct computations). In general I don't know. Edit In general it remains true. As we are working with a normal variety and invertible sheaves, the global sections on a open subset depend only on points of codimension 1, this means we can restrict ourselve to the case $U$ is complement of a divisor $D$. We can of course take $D>0$. If $D$ projects surjective onto the two factors $\mathbb P^n, \mathbb P^m$, then $D$ is ample, so $U$ is affine and we are done. Otherwise $D$ is union of $E_i\times\mathbb P^m$ and of $\mathbb P^n\times E'_j$, then $U$ is a product and we win again.
3) If $X=\mathbb P^n\times \mathbb P^m$ and $F=\pi_1^*L_1$, $G=\pi_2^*L_2$, then use Künneth formula. Otherwise I don't know, but this should be a separate question. 
