There is no "purpose" for the condition that the number of vertices be divisible by 3 – it is part of the problem statement. DHAM3 is the special case of DHAM in which the number of vertices is divisible by 3. In principle this might make the problem easier (a special case can only be easier), but in this case DHAM3 is also NP-hard (and so NP-complete), as you can show by reduction from DHAM.
The problem SHAM3 is not a decision problem and so not eligible to be NP-complete or NP-hard.
How do you show that DHAM3 is NP-hard by reduction from DHAM? You have to come up with an efficiently computable function $f$ that takes an instance $x$ of DHAM and outputs an instance $f(x)$ of DHAM3 in such a way that $x \in DHAM$ iff $f(x) \in DHAM3$. In other words, $f$ takes a graph $G$ and outputs a graph $f(G)$ whose number of vertices is divisible by 3, and moreover $G$ is Hamiltonian iff $f(G)$ is Hamiltonian.
The difficulty here is that the original graph $G$ might have a number of vertices which is not divisible by 3 (otherwise you can take $f(G) = G$). I'll let you figure out how to overcome this difficulty.