As pointed out in my comment, an example would be given by the counting measures on $\mathbb Q \cap [0,1]$ and $(\sqrt{2} + \mathbb Q)\cap [0,1]$, respectively, on the compact metric space $X = [0,1]$.
Note that the same idea actually works for any compact metric space $X$ which has no isolated points.
Since you also asked about an example where the measure spaces are finite:
You can simply take "weighted measures", i.e. if $\{q_n\}_{n \in \mathbb N}$ is a enumeration of $\mathbb Q$, define a function
$$f(x) = \begin{cases} 2^{-n} & \text{if } x = q_n \\ 0 & \text{otherwise} \end{cases} $$
Now the weighted measure is given by $d\tilde \mu = f \, d\mu$, where $\mu$ is the counting measure on the rationals. This will then be finite
$$\int_\mathbb{R} \; d\tilde\mu = \int_\mathbb{R} f \; d\mu = \sum_{n = 1}^\infty 2^{-n} = 1$$
I am not completely sure, and I cannot provide a publicly available reference, but I read in some lecture notes from our university that this decomposition can be generalized to $\mathbb{R}^n$. Let $\lambda$ be the Lebesgue measure on $\mathbb{R}^n$ and $\mu$ the measure under consideration. Then,
$$
\mu = \mu_a + \mu_s + \mu_d
$$
where $\mu_d$ is discrete (i.e., supported on a countable set, with positive measure for every atom), $\mu_a$ is absolutely continuous w.r.t. $\lambda$ (i.e., it possesses a density), and $\mu_s$ is singularly continuous, i.e., it is supported on a Lebesgue null-set, and the atoms of this set have zero measure.
An example for $\mu_s$ in $\mathbb{R}^2$ would be a measure which is supported on a one-dimensional submanifold of $\mathbb{R}^2$, e.g., the uniform distribution on the unit circle.
Best Answer
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 $\mu$.