Theses are simple and natural questions, but I could not find anything about it. If anyone has an answer or a reference this would be very much appreciated.


Let $\mathcal{C}$ be an abelian category (possibly without enough injective nor projective).


(i) Let $A,B \in \mathcal{C}$. When are the $\mathrm{Ext}^n(A,B)$ (defined using Yoneda extensions) sets ?


(ii) Let $A \in \mathcal{C}$ and suppose that $\mathrm{Ext}^n(A,B)$ is a set for all $B \in \mathcal{C}$. Is $\mathrm{Ext}^\bullet(A,-)$ a $\delta$-functor ? If yes is it universal ?


(iii) Same as (ii) in the special case where $\mathcal{C}$ has enough projective.


PS : I edited a bit the question in view of Fernando Muro's comments.



Universality of Ext functor using Yoneda extensions
In their paper entitled "Extension categories and their homotopy", Neeman and Retakh define a spectrum of extensions $\operatorname{Ext}(A,B)$ for any two objects in an exact category $\mathcal E$ such that $\pi\_{-n}\operatorname{Ext}(A,B)=\operatorname{Ext}\_{\mathcal E}^n(A,B)$, in the sense of Yoneda, for any $n\geq 0$. Positive-dimensional homotopy groups vanish. The spectrum $\operatorname{Ext}(A,B)$ is an $\Omega$-spectrum defined by the classifying spaces of the categories $\operatorname{Ext}^n(A,B)$ of $n$-fold Yoneda extensions. 


Given a short exact sequence $B\hookrightarrow C\twoheadrightarrow D$, 
Quillen's Theorem B shows that the homotopy fiber of $\operatorname{Ext}^n(A,C)\rightarrow \operatorname{Ext}^n(A,D)$ is $\operatorname{Ext}^n(A,B)$, $n\geq 0$. Hence, for spectra, the homotopy fiber of $\operatorname{Ext}(A,C)\rightarrow \operatorname{Ext}(A,D)$ is $\operatorname{Ext}(A,B)$. The long exact sequence on homotopy groups defines now a $\delta$-functor $\operatorname{Ext}^\bullet(A,-)$.


Universality follows from Yoneda's lemma. If $T$ is another $\delta$-functor, a natural transformation $\operatorname{Hom}(A,-)=\operatorname{Ext}^0(A,-)\rightarrow T^0$ extends uniquely to a morphism of $\delta$-functors $\operatorname{Ext}^n(A,-)\rightarrow T^n$, $n\geq 0$, as follows. An $n$-fold extension $B\hookrightarrow X\_1\rightarrow\cdots\rightarrow X\_n\twoheadrightarrow A$ factors as the 'composition' of short exact sequences
$$Y\_{n-1}\hookrightarrow X\_n\twoheadrightarrow Y\_n$$
with $Y\_0=B$ and $Y\_n=A$. In particular we obtain morphisms
$$T^0(A)\rightarrow T^1(Y\_{n-1})\rightarrow T^2(Y\_{n-2})\rightarrow\cdots\rightarrow T^{n-1}(Y\_1)\rightarrow T^n(B).$$
The image of the previous extension by extension by $\operatorname{Ext}^n(A,B)\rightarrow T^n(B)$ is the image by this composite of the element in $T^0(A)$ classifying the natural transformation we started with (via Yoneda's lemma). Everything is well defined by the properties defining a $\delta$-functor.


This is only a sketch of proof. If you intend to use it in a paper you should probably provide some details at some points, e.g. carefully check the hypotheses of Quillen's Theorem B.


