$k$-coloring in BPP, implies $k$-coloring in ZPP
This statement is to my knowledge unknown. If this is an exercise, then it is likely an error: did they mean $RP$ instead of $ZPP$?

Since $k$-coloring is NP-complete, what you are asked to show is:


  If $NP \subseteq BPP$, then $NP = ZPP$.


First, let's review what is known: the basic inclusions are the following:
\begin{align*}
P \subseteq ZPP \subseteq RP &amp;\subseteq NP \\
P \subseteq ZPP \subseteq coRP &amp;\subseteq coNP \\
                          RP &amp;\subseteq BPP \\
                          coRP &amp;\subseteq BPP.
\end{align*}

That is, $P$ is contained in $ZPP = RP \cap coRP$, which is contained in $BPP$. Moreover, $RP$ is contained in $NP$ and $coRP$ is contained in $coNP$.
But we don't know how $NP$ and $coNP$ relate to $BPP$ (it is conjectured that $P = ZPP = BPP$, which implies that $NP$ and $coNP$ contain $BPP$).

Now, what happens if (as in your exercise) $NP \subseteq BPP$? It follows (and is a common exercise to show) that
$$
NP = RP.
$$

(see e.g. here).
That means that
$$
coRP = coNP \\
ZPP = NP \cap coNP \\
$$
So basically in this case, we have $P$, which contains $NP$ and $coNP$, and those are both contained in $BPP$ (which also equals the polynomial hierarchy $PH$).

But there is no reason to believe that this implies that $NP = coNP$, which is what you would need to then show that your $NP$-complete language, $k$-col, is in $ZPP = NP \cap coNP$. In particular, we can imagine that $BPP$ is equal to the second level of the polynomial hierarchy (you don't need to be familiar with this, but it's just some level that contains both $NP$ and $coNP$). That would imply that $NP$ is contained in $BPP$, but it doesn't mean that $NP$ and $coNP$ are equal.

TL;DR The statement you are trying to show appears to be unknown.