If $\tau_1$ and $\tau_2$ are topologies on the same set $X$, we say that $\tau_1$ is smaller or coarser than $\tau_2$ if $\tau_1\subseteq\tau_2$. (Note that this terminology is a little sloppy, since $\tau_1$ is allowed to be equal to $\tau_2$. If you want to specify that $\tau_1\subsetneqq\tau_2$, it’s best to say that $\tau_1$ is strictly weaker or coarser than $\tau_2$.
For a simple example, let $\tau_1$ be the usual Euclidean topology on $\mathbb{R}$, and let $\tau_2$ be the discrete topology on $\mathbb{R}$; since $\tau_2=\wp(\mathbb{R})$, it’s clear that $\tau_1\subseteq \tau_2$, so the Euclidean topology on $\mathbb{R}$ is weaker (or coarser) than the discrete topology. (In fact it’s obviously strictly weaker.) On the other hand, if we let $\tau_3=\{\mathbb{R}\setminus F:F\text{ is finite}\}$, the cofinite topology on $\mathbb{R}$, then $\tau_3\subsetneqq\tau_1$: the cofinite topology is strictly weaker than the Euclidean topology.
Added: A useful fact is that if $\mathscr{S}$ is any collection of subsets of $X$, and $\tau$ is the intersection of all of the topologies on $X$ that contain $\mathscr{S}$, then not only is $\mathscr{S}\subseteq\tau$, but $\tau$ is again a topology on $X$. (This is an easy exercise using the definition of a topology.) Clearly, then, $\tau$ is the weakest (smallest, coarsest) topology on $X$ that contains every member of $\mathscr{S}$, which therefore always exists.
This means that if you have some collection $\mathscr{S}$ of sets that you want to be open, you can find the weakest (smallest, coarsest) topology that does the trick by intersecting all of the topologies that make the members of $\mathscr{S}$ open. The collection $\mathscr{S}$ is said to be a subbase for the resulting topology.
To return to the specifics of your question, notice that the more restrictions you put on $f$, the more open sets you’ll require in $X$. For instance, $\tau_5$ makes $f$ both open and closed, so it automatically makes $f$ open. Thus, it’s one of the topologies that are intersected to find $\tau_3$, and as a result $\tau_3\subseteq\tau_5$.