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
$$\int_\mathbb{R} \; d\tilde\mu = \int_\mathbb{R} f \; d\mu = \sum_{n = 1}^\infty 2^{-n} = 1$$
Untrue. Consider $\Omega=\mathbb R^d$, $\mathcal F=\mathcal B(\mathbb R^d)$, $X$ the identity, and $\mathbb P$ and $\mathbb Q$ some discrete probability measures with positive weights at the elements of some countable sets $S$ and $T$, respectively.
Then the supports of $\mathbb P=\mu_\mathbb P$ and $\mathbb Q=\mu_\mathbb Q$ are the closures of $S$ and $T$. These may coincide while $S\cap T$ is empty, that is, one can have $\mathrm{supp}(\mu_\mathbb P)=\mathrm{supp}(\mu_\mathbb Q)$ while $\mathbb P$ and $\mathbb Q$ are mutually singular.
Example: if $d=1$, one can choose $S$ the set of rational numbers whose reduced form has an even denominator and $T$ the set of rational numbers whose reduced form has an odd denominator.
Best Answer
This question is kind of vague. In some sense the answer is "nothing". As pointed out in the comments, a sequence of absolutely continuous measures can converge to a point mass, and a sequence of purely atomic measures can converge to an absolutely continuous one. You can also have all of the supports be disjoint from each other ($\mu_n = $ a point mass at $1/n$).