Under what conditions $(X,\tau)$ accepts a finite measure? Under what condition we can guarantee that a topological space $(X,\tau)$ accept a finite measure?
For example if the space $X$ is compact or compact Housdorff then can we find a finite measure on $X$?
 A: You can always get a finite measure on $X$ by choosing your favourite $p \in X$ and defining $\mu(E) = 1 \iff p \in E$ and $\mu(E)=0$ otherwise. 
Note this measure has nothing to do with the topology on $X$ since $\mu$ is defined for both Borel and non-Borel subsets. 
In case this measure is not the sort of thing you're looking for, you should put some restrictions on the measure. For example $\mu(U) > 0$ for all open $U \subset X$. One type of space this rules out is anything with a family of uncountably many pairwise disjoint open sets. See this answer.

Edit: For compact metric $X$ we can define the Hausdorff measure.
$\displaystyle \mu_\delta(E) = \Big \{ \sum \text{diam} \ U_n \ \big \vert  \ E \subset \bigcup U_n \text{ and all diam} U_n < \delta\Big \}$
$\mu(E) = \inf \big \{\displaystyle \mu_\delta(E) : \delta > 0 \big \}$.
It shouldn't be too hard to show each open set has nonzero measure, and the measure of the whole space can be bounded above by some formula involving the diameter of $X$ which is finite by compactness.
For $X$ not metric but merely Hausdorff we cannot do this. For example consider the one-point compactification $D^*$ of a discrete space $A$ of uncountable cardinality. In case each $\mu( \{a\}) > 0$ a similar argument to the linked answer entails $\mu(D^*) = \infty$.
A: Finite measures always exist, on any set. Most of the time you at least want the Borel sets to be measurable to have any topology-measure connection.
There is a sizable theory on topological measure theory (see Fremlin's book, part 4 of 5 volumes on measure theory in general..) about such subjects. There is a lot of interaction between measures of certain types ((inner/outer) regular, strictly positive etc.) and several topological properties. 
So e.g. if you want a strictly positive Borel ($\sigma$)-finite measure (i.e. $\mu(U) > 0$ for all open non-empty $U$), this means that your space must be ccc (every family of pairwise disjoint non-empty open sets is at most countable), but I don't think every ccc space necessarily admits such a measure, but a separable space does (take a weighted sums of point measures on a dense subset). 
Topological groups have a Haar measure (not always finite, I think compactness is enough to get finiteness though), etc.  
