How will I know which quantum numbers are the lowest energy?
In general that is a difficult question from first principles. Simulation can often answer it, but the problem can be pretty involved and demanding.
However, as a practical matter the configuration and energy levels of many nuclei are known from extensive experiments. For example an online level diagram and a table of level data for Gd-157 from http://www.nndc.bnl.gov/nudat2/ .
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
A better reference for determining these nuclear states from experiment is NuDat at Brookhaven. Figuring out the interaction energy of a neutron in any particular energy level with a $^{16}O$ core is extremely difficult. There is some model intuition that can help you, as you used for the first excited state. In this case, the second excited state $1/2^-$ must be more complicated. I would guess that it results from a neutron from the closed core joining the extra neutron in the d-shell, leaving a hole in the p-shell that gives the quantum numbers here. As you can see, there's a huge extra energy needed to make this configuration.