$\mathcal{T}_1:=\{A \subset X : A = \emptyset \lor |A^c| < \infty\}$ Topology? Let $X \neq \emptyset$. And $(.)^c$ is the complement.
$\mathcal{T}_1:=\{A \subset X : A = \emptyset \lor |A^c| <  \infty\}$
So for my first case I looked at $|X|<\infty$.
Then $X \notin \mathcal{T}$ because $|X| \neq \emptyset$ and $|X^c|$ might be $\not< \infty$.
So $\mathcal{T_1}$ can't be a topology for finite $X$.
For $|X|=\infty$:
Then the complement of $X$ can also be infinite so this can't be a topology either because $X \notin \mathcal{T_1}$.
$\mathcal{T_2}:=\{A \subset X: A=X \lor |A|<\infty\}$
$\emptyset, X \in \mathcal{T_2}$ 
$A_1 \cap A_2 \in \mathcal{T_2}$
$\bigcup_{i \in I}A_i \in \mathcal{T_2}$? I am not sure about this one
 A: Case 1: $|X|$ is finite.
This is almost trivial as every subset of $X$ will be in $\mathcal{T}_1$ (as every set will have a finite complement), so you wind up with the discrete topology, which is a topology.
Case 2: $|X|$ is infinite.
If you have two subsets such that their complements are finite, then their union will have a complement that is the intersection of their individual complements. The intersection of finite sets is finite, so the union will still be in $\mathcal{T}_1$. Additionally, the full set and the empty set are obviously there. Next, check finite intersections. The complement of finite intersections is the union of the complements (or at least a subset of it). The finite union of finite sets is finite, so it will still be in $\mathcal{T}_1$.
Therefore, $\mathcal{T}_1$ is, indeed, a topology.  
Really, they are the same case.
Edit: For $\mathcal{T}_2$, it is not a topology. Example: Let $X=\mathbb{N}$. Then the sets of $\mathcal{T}_2$ will be all finite sets of natural numbers (and the full set). The following union of finite sets is an infinite set that is not the full set (and therefore not in $\mathcal{T}_2$):
$$\displaystyle \bigcup_{n=2}^\infty \{n\}$$
Since $\mathcal{T}_2$ does not contain all unions, it is not a topology.
