The Kazdan-Warner theorem goes a long way toward answering the first and second questions.
(For notes typed up by Kazdan, see http://www.math.upenn.edu/~kazdan/japan/japan.pdf.)
Here's what is says (taken almost verbatim from the notes, page 93): Divide the class of all closed manifolds (edit: of dimension > 2. See comments) into 3 types:
I. Those which admit a metric of nonnegative scalar curvature which is positive somewhere.
II. Those which don't but admit a metric of 0 scalar curvature.
III. All other closed manifolds.
The theorem is that if $M$ is in class I, then any $f:M\rightarrow\mathbb{R}$ is the scalar curvature of some metric.
If $M$ is in class II, then $f:M\rightarrow\mathbb{R}$ is the scalar curvature of some metric iff it's identically 0 or negative somewhere.
If M is in class III, then $f:M\rightarrow\mathbb{R}$ is the scalar curvature of some metric iff it's negative somewhere.
In particular, every closed manifold has a metric of constant negative scalar curvature. Those in class I or II have a metric of 0 scalar curvature, and those in class I have a metric of constant positive scalar curvature.
The simplest example is $S^n$, it is locally conformally flat with the standard metric,
and is not flat for obvious reasons.
While flat manifolds are precisely quotients of $\mathbb R^n$ by discreet group of isometries, one should not expect to have a classification of conformally flat manifolds in higher dimensions. For example, already in dimension $4$ it was proven by Kapovich in
M. Kapovich. Conformally flat metrics on 4-manifolds. J. Differential
Geom. 66 (2004), no. 2, 289–301,
that arbitrary finitely presented group can be a subgroup of a fundamental group of a conformally flat manifold.
The article of Kapovich is and from its introduction you will learn a lot on the question. $4$-dimensional manifolds with LCF structure have zero signature, in dimension $3$ it is known that some manfiolds don't admit conformally flat structure, first example was constructed in W. Goldman, Conformally flat manifolds with nilpotent holonomy and the uniformization problem for $3$-manifolds, Transactions
of AMS 278 (1983).
One more remark -- all hyperbolic manifolds (of constant negative sectional curvature) are all conformally flat. A connected sum of two conformally flat manifolds is conformally flat and so this already gives you a large collection of examples.
Best Answer
Take a unit sphere $S^2$ and a hyperbolic surface $X$. Then the product $S^ 2 \times X$ is not flat and has zero scalar curvature. Also it is conformally flat by a paper
Simon Salamon (2009) Complex structures and conformal geometry. In: BOLLETTINO DELLA UNIONE MATEMATICA ITALIANA, vol. 9, pp. 199-224
as per editor Holonomia.