What would happen if Hofmann bromamide reaction is carried out in Br2 and KOD?
This isn't the most well-thought-out scenario, because if you dissolve $\ce{KOD}$ in $\ce{H2O}$ all the deuteriums will be lost immediately anyway.

$$\ce{KOD + H2O &lt;=&gt;&gt; KOH + HOD}$$

Assuming that the amount of $\ce{KOD}$ is much less than the amount of $\ce{H2O}$ (after all $\ce{H2O}$ is the solvent) then the position of this equilibrium will lie very, very far to the right. So, at the end of the day, if you are using $\ce{H2O}$ as the solvent, the use of $\ce{KOD}$ won't have any effect. If you want to see any difference in deuteration, you have to at least use $\ce{KOD/D2O}$. However, there is a second problem with this:

If you dissolve an amine in $\ce{D2O}$, then all the amine hydrogens will be very rapidly replaced with deuteriums. It does not even matter mechanistically what the pathway is. Regardless of whether the initial product is $\ce{RNH2}$, $\ce{RNHD}$, or $\ce{RND2}$, as long as there is $\ce{D2O}$ in the mixture, all three will become $\ce{RND2}$. The same can be said of dissolving them in $\ce{H2O}$.



To answer your questions:

(1) No. In general, protonation can and will occur by any solvent molecule that happens to be close by. If you are running the reaction in $\ce{D2O}$ solvent, then there is much, much more $\ce{D2O}$ than there will ever be $\ce{H2O}$ produced. So, protonation by $\ce{H2O}$ is statistically close to impossible.

(2) None of us can answer that, as we cannot presume what your teacher was thinking of when they set the question.

(3) Depends on what your solvent is, see comments above on dissolving amines in $\ce{D2O}$ or $\ce{H2O}$.