The number of neutrons is even, so it indeed means that they contribute spin zero and positive parity.
The spin and parity comes from the "last proton" because the number of protons is odd.
The dependence of the energy on the angular momentum is such that the pairs at a high value of $J$ are preferred (lower in energy) due to the special, spin-dependent features of the strong nuclear force (features invisible in the single-nucleon model). That's true despite the fact that the single-particle shells with a lower $J$ could be preferred.
It follows that among the 3 protons in $1d_{5/2}$, the pair really chooses $j_z=\pm 5/2$, the maximum value (in the absolute value). The remaining slots $j_z=\pm 1/2$ and $\pm 3/2$ are available for the last proton. The last proton also prefers the higher value of $J$ so it will sit in the $J=3/2$ state. It's a $d$-shell, i.e. $l=2$, so the parity is $(-1)^l=+1$.
Computing the excitation spectrum for a nucleus — that is, its energy levels and their quantum numbers — is hard. Consider that the Schrödinger equation has no known exact solutions for atoms other than hydrogen: a helium atom, with three charges instead of two, is too complex to be treated except in approximation. The nuclear many-body problem is much thornier. Not only are there more participants in the system (ninety-five, for $^{95}$Nb), but in addition to the electrical interaction among the protons you have the pion-mediated attractive strong force, the rho- and omega-mediated hard-core repulsion, three-body forces, etc. (I'm impressed that you got the correct spins and parities for the ground states from the shell model; nice work!)
So when normal people want to know the spin and parity and energy of a nuclear state, we look it up. The best source is the National Nuclear Data Center hosted by Brookhaven National Lab, which maintains several different databases of nuclear data (each with its own steep learning curve). Searching the Evaluated Nuclear Structure Data File by decay for $^{95}$Nb brings up level schemes, with references and lots of ancillary data, for both niobium and molybdenum. These confirm that you've gotten the $J^P$ for the ground states correct. Two excited states are listed for $^{95}$Mo: one at $200\rm\, keV$ with $J^P = 3/2^+$, and the one you mention at $766\rm\,keV$ with $J^P=7/2^+$. You can follow the references to see the experimental arguments for assigning those spins and parities.
You can make some general, hand-waving predictions about spins by thinking about angular momentum conservation in the transitions.
The matrix element for a particular transition is generally proportional to the overlap between the initial wavefunction and the final wavefunction.
In nuclear decays the initial state is the nucleus, which is tiny and more-or-less spherical with uniform density, while the final state includes the daughter nucleus and the wavefunctions for the decay products. If the decay products carry orbital angular momentum $\ell$, the radial part of the wavefunction goes like $r^\ell$ near the origin. Dimensional analysis then says that the overlap between the nucleus and the decay wavefunction is proportional to $(kR)^\ell$, where $R$ is the nuclear radius and $k = p/\hbar = 2\pi/\lambda$ is the wavenumber of the decay product. (Note that nuclear decay products typically have $\lambda \gg R$, so you can treat the decay product wavefunction as roughly uniform averaged over the nucleus.)
If the probability of a decay is proportional to $(kR)^\ell$, that means that
decays where the product's momentum $p=\hbar k$ is large are preferred over decays where the product's momentum is small
decays where the orbital angular momentum $\ell$ is small are preferred over decays where $\ell$ is large
For your $\rm Nb\to Mo$ transition, the decay to the excited state is preferred over the $\frac92\to\frac52$ ground-state-to-ground-state transition, which suggests that the excited state probably has spin $\frac72, \frac92, \frac{11}2$. The most probable of these is $\frac72$, since angular momentum tends to relax during decay processes — and indeed, that's the spin of the $766\rm\,keV$ excited state.
Best Answer
The $d$ states have orbital angular momentum $L=2$. The parity is related to orbital angular momentum, not total angular momentum $j=5/2$.
The previous closed shell contains $8$ neutrons. Given it’s closed, you can deduce the value of total spin $S$ for the first $8$ neutrons. Since neutrons have $s=1/2$, it’s not hard to figure out what are the possible values of $S_{tot}$ for the system of $8+1$ neutrons. As a final check you can verify that $j=5/2$ is indeed a possible in the coupling of this $S_{tot}$ to orbital angular momentum $L$.