Can we recover a topology from knowing about continuity of maps from compact spaces? Explicitly:
Question. Suppose $\alpha$ and $\beta$ are topologies on a common set $X$ such that for all compact Hausdorff spaces $K$ and all function $f : K \rightarrow X$, it holds that $f$ is $\alpha$-continuous iff $f$ is $\beta$-continuous. Is it true that $\alpha = \beta$?
Follow up question: If the answer is "no", can this be solved either by dropping the assumption that $K$ be Hausdorff, or else by assuming that $X$ is Hausdorff?
The answer is no in all your cases. For instance, let $S$ be an uncountable set, let $X=S\cup\{\infty\}$, let $\beta$ be the discrete topology, and let $\alpha$ be the topology in which every subset of $S$ is open and a set containing $\infty$ is open iff it is cocountable. These two Hausdorff topologies have the exact same continuous maps from compact spaces, since any compact subset in either topology is finite and the two topologies agree on finite sets.
More generally, given a topology $\alpha$ on $X$, you can let $\beta$ be the finest topology on $X$ with respect to which all $\alpha$-continuous maps from compact Hausdorff spaces are still continuous (in other words, $\beta$ consists of all $U\subseteq X$ such that $f^{-1}(U)$ is open for every $\alpha$-continuous map $f:K\to X$ with $K$ compact Hausdorff). If $\alpha\neq\beta$, this will give a counterexample. If $\alpha=\beta$, then the topology $\alpha$ is called compactly generated. So, the answer to your question would be yes (essentially by definition) if you require $\alpha$ and $\beta$ to be compactly generated. Compactly generated topologies are a pretty broad class: they include all first-countable topologies and all locally compact Hausdorff topologies, for instance.