No way: Think of Lebesgue measure and a point measure.
You can often speak of the support up to a null-set, but in order to single out a specific support you need further structure on your measure space.
Added: In 2. I was deliberately a bit sloppy. A sufficient condition for the existence of a good notion of support is a class of null sets closed under arbitrary unions. The union of those null-sets is then the largest such $\mu$-null set and its complement deserves the name of support of $\mu$. For instance, if $\mu$ happens to be a Radon measure on a locally compact space, you can take the class of open $\mu$-null sets and the union of those is precisely the complement of the (closed) support of $\mu$.
I will concentrate on comparing (3) and (4). The definition (1) is meant for finite signed measures, whereas all the other definitions are meant for arbitrary positive measures; (1) is equivalent to (4) in the case of finite positive measures. (2) appears to be equivalent to (4) ["locally finite" can mean "finite on compact sets", although it is sometimes taken to mean "finite on the elements of some topological basis"; these are equivalent in the LCH (locally compact Hausdorff) case]. Finally, (5) does not appear to be a definition at all, but rather a description of a definition.
Now then,
i) In the case of a second countable LCH space, every locally finite measure satisfies both (3) and (4) (Theorem 7.8 of [1]). This is the most commonly considered scenario in applications, which is why almost no one bothers to carefully sort out the differences between the different definitions.
ii) In the case of a sigma-compact LCH space, (3) and (4) are equivalent. The forward direction is Corollary 7.6 of [1]; the backwards direction follows from the forward direction together with (iv) below (but I'm sure there is an easier proof).
iii) (3) and (4) are not equivalent in general, even for LCH metrizable spaces (Exercise 7.12 of [1]).
iv) In an LCH space, there is a bijection between
A) measures satisfying (3),
B) measures satisfying (4), and
C) positive linear functionals on the space of continuous functions with compact support.
(The Riesz representation theorem gives either (A)<->(C) or (B)<->(C), depending on where you look; (A)<->(B) is in the Schwarz book mentioned by Joe Lucke; see also Exercise 7.14 of [1])
[1] G. B. Folland, Real Analysis: Modern Techniques and Their Applications
Note: In [1], "Radon" refers to measures satisfying (3).
Best Answer
For an example with a probability measure, consider the following standard counterexample: let $X = [0, \omega_1]$ be the uncountable ordinal space (with endpoint), with its order topology. This is a compact Hausdorff space which is not metrizable. Define a probability measure on the Borel sets of $X$ by taking $\mu(B) = 1$ if $B$ contains a closed uncountable set, $\mu(B)=0$ otherwise. It is known that this defines a countably additive probability measure; see Example 7.1.3 of Bogachev's Measure Theory for the details.
If $x \in X$ and $x < \omega_1$, then $[0, x+1)$ is an open neighborhood of $x$ which is countable, hence has measure zero. So $x$ is not in the support of $\mu$. In fact the support of $\mu$ is simply $\{\omega_1\}$. But $\mu(\{\omega_1\}) = 0$.