If two topological spaces have the same topological properties, are they homeomorphic?

Well, it usually goes the other way. A property $P$ of a topological space $X$ is deserved to be called topological if $P(X)$ holds if and only if $P(Y)$ holds whenever $X$ and $Y$ are homeomorphic. An example of a topological property is "$X$ is connected" while an example of a non-topological property is "$X$ is a subset of $\mathbb{R}^n$. The latter can be upgraded to a topological property by requiring instead "$X$ can be embedded in $\mathbb{R}^n$".

With this convention, given a topological space $X$ there is a smart-ass topological property one can define using $X$: "$P(Z)$ holds iff $Z$ is homeomorphic to $X$". Since homeomorphism is an equivalence relation, this is indeed a topological property and clearly $X$ satisfies $P$. If $Y$ is another space that satisfies the same topological properties as $X$, then $P(Y)$ must hold and so $X$ and $Y$ are homeomorphic.

Yes.

One topological property of $X$ is belonging to the homeomorphism (equivalence) class of $X$.

For two arbitrary topological spaces $X$ and $Y$, there isn't a nice list of things you can check off on each of them to be able to say they are homeomorphic. The only thing you can really do to show $X$ and $Y$ are homeomorphic is demonstrate the existence of a homeomorphism.

On the other hand, there are certain classes of topological spaces for which there exist complete classifications. For example, a one dimensional, simply connected complex manifold $X$ is homeomorphic to either the complex plane or its one point compactification, and this is determined by whether or not $X$ is compact.