Given two smooth projective schemes $X$ and $Y$ over some algebraically closed field $k$, we have $X\times Y$ with the projections $p$ to $X$ and $q$ to $Y$. Furthermore we have a "nice" sheaf of algebras $R$ on $X$, i.e. locally free and of global dimension at most dim($X$), e.g. Azumaya or something similar. Given two $p^{\*}R$$p^{*}R$-modules $M$ and $N$ on $X\times Y$, which are coherent and torsion free.
Like in the commutative case, i define the i-th relative $\mathcal{E}xt$-sheaf on $Y$ to be: $\mathcal{E}xt^i_{p^{\*}R,q}(M,N):=(R^i(q_{\*}\mathcal{H}om_{p^{\*}R}(M,-))(N)$$\mathcal{E}xt^i_{p^{*}R,q}(M,N):=(R^i(q_{*}\mathcal{H}om_{p^{*}R}(M,-))(N)$
Can I expect them to have the same properties as in the commutative case?
For example:
(1) Do we have $\mathcal{E}xt^i_{p^{\*}R,q}(M,N)=0$$\mathcal{E}xt^i_{p^{*}R,q}(M,N)=0$ for $i>dim(X)$?
(2) Given $y\in Y$ is there a map $\mathcal{E}xt^i_{p^{\*}R,q}(M,N)\otimes k(y) \rightarrow Ext_R^i(M_y,N_y)$$\mathcal{E}xt^i_{p^{*}R,q}(M,N)\otimes k(y) \rightarrow Ext_R^i(M_y,N_y)$
(3) If $Ext_R^i(M_y,N_y)=0$ for all $y\in Y$ does this imply $\mathcal{E}xt^i_{p^{\*}R,q}(M,N)=0$$\mathcal{E}xt^i_{p^{*}R,q}(M,N)=0$?
(4) Is there a kind of base change theorem for the $\mathcal{E}xt^i_{p^{\*}R,q}(M,N)$$\mathcal{E}xt^i_{p^{*}R,q}(M,N)$?
Or do I need more conditions for $M$ and $N$ to have the desired properties? I'm especially interested in the case, where $M=p^{\*}P$$M=p^{*}P$ for some $R$-module $P$ on $X$.
