The only additional constraint which is required for $h_3$ and $h_4$ to be admissible is that both $h_1(n)$ and $h_2(n)$ return non-negative values for any state $n$, but this is always accomplished ---even if a heuristic function might return a negative value it just suffices to take the maximum between the value returned and 0, and this operation clearly preserves admissibility.
Thus, given that $h_1(n), h_2(n) \geq 0$ for every state $n$ and that both are admissible, it is easy to see that the values of $h_3(n)$ are upper bounded by those of $h_1(n)$ (the maximum value taking place when $h_2(n)=0$) Since $h_1$ is admissible so it is $h_3$; conversely, if $h_2(n)$ returns any strictly positive value then $h_3(n) < h_1(n)$. Now, if $h_1(n)$ is not overestimating the effort to reach the goal, then $h_3(n)$ is neither as well and thus, $h_3$ is admissible in this case also.
The proof of admissibility of $h_4$ follows the same lines.
Note that in these proofs it is never required for any heuristic function to strictly dominate another, i.e., the condition $h_1(n) < h_2(n)$ is useless and not necessary at all. The only important requirement is that none ever return negative values, but this is a trivial requirement.
To conclude, I think this is mostly a theoretical exercise. Note that as mentioned above $h_3$ and $h_4$ are upper bounded by $h_1$ and $h_2$ respectively so that, in other words, they are less informed than them. In other words, these transformations are of no interest.
Hope this helps,