What is an example of a group $G$ which 


1- is finitely generated by $S$,


2- does not have property (T), 


3- admits infinitely many finite quotients which do not factor through an homomorphism $G \to H$ for some [fixed & infinite] property (T) group $H$


4- all the Cayley graphs (w.r.t. $S$) of those finite quotients are $\epsilon$-expanders (for some fixed $\epsilon >0$)


The question is motivated by the *false* assertion: "a group has property (T) if and only all its finite quotients are $\epsilon$-expanders".


There are "simple" answers if one does not put the factor condition in 3. For example, one could consider $A \times H$ where $A$ is a finitely generated simple amenable group (as proven/constructed by Juschenko-Monod/Matui) and $H$ is some residually finite property (T) group ($SL\_3(\mathbb{Z})$ being the canonical choice). Because $A$ is amenable, this does not have (T) and its finite quotients (which all come from $H$) are $\epsilon$ expanders. 


Actually, to enlarge the list of examples of this form, a side question would be: what are the finitely generated groups which do not have property (T) and do not have finite non-trivial quotients?



Groups without property (T) but all finite quotients are expanders
I think your condition (4) is called "property ($\tau$)". (See Theorem 4.3.2 of Lubotzky's very nice book "Discrete Groups, Expanding Graphs and Invariant Measures".) An example is $G = \mathrm{SL}\_2\bigl(\mathbb{Z}[1/p] \bigr)$. (See Example 4.3.3E on page 52 of Lubotzky's book.)


