Your question is in general how to compute the exact microcanonical ensemble, since your degeneracy $g$ is the microcanonical partition function $W$. The usual strategy is to compute the canonical ensemble which has a simpler form and go back to the microcanonical one. Technically, the canonical ensemble partition function $Z(\beta)$ is the Laplace transform of $W(E)$, so you can go back by inverse Laplace transform. In your case, the energy spectrum has integer values in appropriate energy units, so a $z$ transform is more natural. Concretely, you have a sequence $W_n$ indexed by energy and $Z$ is associated generating function, i.e. the integer power series in $x = e^{-\beta}$:
$$
Z = \sum_{m=0}^\infty W_mx^m
$$
Take for example the case $N=1$. Setting the ground state to be zero, the partition function factorizes. From the microcanoncial perspective, the total degeneracy is the convolution of the individual degeneracy by independence, but the transform converts it to the usual product:
$$
Z_1 = \frac1{(1-x)^D}
$$
so by the binomial theorem:
$$
W_{m,1} = (-1)^m\binom{-D}m = \binom{D+m}m
$$
which is consistent with your multiplet interpretation.
In the case of $N$ independent quantum harmonic oscillators, the canonical ensemble partition function again factorizes (by the same argument), so:
$$
Z = Z_1^N =\frac1{(1-x)^{DN}}
$$
and you get:
$$
W_m = (-1)^m\binom{-DN}m=\binom{DN+m}m
$$
which you can also obtain by elementary combinatorial arguments. This gives the degeneracy of $N$ independent harmonic oscillators.
For i), I guess that you are asking about the degeneracy of $N$ distinguishable particles distributed over the orbitals of an $D$ dimensional harmonic oscillator. In this case, the partition function is just:
$$
Z = \frac1{N!}Z_1^N
$$
with the factorial prefactor there for Gibbs' paradox and is justified by Maxwell-Boltzmann statistics. This gives you:
$$
W_{mn} = \frac1{n!}\binom{DN+m}m
$$
Note that to get a proper thermodynamic limit, you'll need to rescale the harmonic trap's strength (analogous to an extensive volume for a gas in a box).
For bosons, even the canonical ensemble is not simple enough, you need to apply the same trick to particle number and upgrade to the grand canonical ensemble which has a simple factorised general form due tot the convolution properties. Again, the grand canonical ensemble is a Laplace transform (up to a different sign convention) $\mathcal Z(\beta\mu)$ of $Z(N)$, but since $N$ has integer values, a $z$ transform is more appropriate, and you should view $\mathcal Z$ as an integer power series of $x$ and $z=e^{\beta\mu}$ (fugacity), whose coefficients give you your desired degeneracy, i.e. the microcanonical ensemble. In short, the grandcanonical function is the generating function:
$$
\mathcal Z = \sum_{m,n=0}^\infty W_{mn}x^mz^n
$$
In general, from Bose-Einstein statistics:
$$
\mathcal Z = \prod_{m=0}^\infty\frac1{(1-zx^m)^{g_m}}
$$
with $g_m$ the degeneracy in the case of a single particle. In your case it is:
$$
g_m = \binom{D+m}m
$$
Your goal now is to expand the product. As usual, you can expand the product directly which will express the $W_{mn}$ as convolutions. In general, I doubt that that there is a simple analytic formula for $W_{mn}$. At best, you have different kinds of expressions, and the only tractable regime is the thermodynamic limit.
Another way to get the expansion is to first calculate $\ln\mathcal Z$. It is sometimes simpler to think in terms of the number of particles in the grancanonical ensemble:
$$
\mathcal N = z\partial_z\ln\mathcal Z
$$
In your case, you can explicitly compute:
$$
\mathcal N = \sum_{n=0}^\infty \frac{z^n}{(1-x^n)^D}
$$
so:
$$
\ln\mathcal Z = \sum_{n=1}^\infty \frac{z^n}{n(1-x^n)^D}
$$
This gives you a new expression:
$$
\mathcal Z = \exp\left(\sum_{n=1}^\infty\frac{z^n}{n(1-x^n)^D}\right)
$$
and expanding it would give you again your degeneracies. This way of rewriting is often more convenient as it makes the thermodynamic limit more accessible and makes the low density correspondence with Maxwell-Boltzmann statistics more evident.
Hope this helps.
