Dualizing complex definition ubiquity The following definition is the one, I found here http://stacks.math.columbia.edu/tag/0A7A. But let me recall it (out of consistency's sake):
Definition
For $A$ a Noetherian ring, a dualizing complex for it is a complex of $A$-modules say $\omega^{\bullet}_A$ with the following properties
1.) $\omega^{\bullet}_A$ has finite injective dimension.
2.) $H^i(\omega^{\bullet}_A)$, is a finite $A$-modules for all $i$.
3.) $A \rightarrow RHom_A(\omega^{\bullet}_A,\omega^{\bullet}_A)$, is a quasi-isomoprhism.
Now many questions come along this definition. Because no other information is provided concerning it, I don't understand various things. 
Firstly, what does it mean for a complex to have finite injective dimension? I think that probably means that since we're working over an abelian category, there is always a (co-chain) map $\omega^{\bullet}_A \rightarrow I^{\bullet}$, which is quasi-isomorphism and the right-hand side complex consisting of injective modules (straightforward from Cartan-Eilenberg resolution). Does finite mean that $I^{\bullet}$ is bounded in that case?
Secondly, concerning 3.), do we treat $A$ as a cochain complex on its own, where we have at $0$ and $1$ position only $A$, and zero elsewhere?
One last question, under which circumstances for the ring $A$ this dualizing complex exists? And why do we substitute the dualizing module with that instead in some cases?
P.S.1
I asked the same question on MSE but I think isn't very easy question for it (even if it is soft in general), so I posted here. 
P.S.2
Of course as soon as I have a response here, will deleted by MSE.
Thank you!
 A: A. Your guess as to what it means for a complex to have finite injective dimension is indeed the standard definition.
B. You treat $A$ as a complex by letting it be concentrated entirely in degree zero.  That is, the zeroth entry is $A$, and all other entries are the zero module.
C. Existence is equivalent to $A$ being the homomorphic image of a Gorenstein ring of finite Krull dimension.  See the mathoverflow question State of the art on a question on the existence of dualizing complex .
D. The 'dualizing module' is the right substitute for dualizing complex in the case where $A$ is Cohen-Macaulay.  What I mean is this: When $A$ is Cohen-Macaulay, then a dualizing module exists if and only if $A$ is the homomorphic image of a Gorenstein ring.  When $A$ is not Cohen-Macaulay, there is no dualizing module.  When a dualizing module exists, its injective resolution satisfies the conditions of a dualizing complex.  Furthermore, any dualizing complex for $A$ is quasi-isomorphic to the dualizing module concentrated in degree zero.  I like the development of dualizing modules (also called 'canonical modules') in the book of Bruns and Herzog, chapter 3.
