= Solution
The separate <proton> and <neutron> shell capacities are $2j+1=2,4,2,6,\ldots$. Filled shells and paired particles have zero net <spin> and <magnetic dipole moment>. The <single-particle nuclear magnetic moment> is $\mu=g_jj\mu_N$, with orbital factors $g_\ell^p=1$, $g_\ell^n=0$ and $s=1/2$.
For nitrogen, the eight <neutrons> form a closed shell. The seventh <proton> occupies $1p_{1/2}$, so $\ell=1$, $j=1/2$ and the <Landé g-factor> gives
$$
g_j=\frac43g_\ell-\frac13g_s=\frac43-\frac{5.586}{3}=-0.52867.
$$
Hence the <nuclear shell model> predicts
$$
\boxed{\mu({}^{15}\mathrm N)=-0.2643\,\mu_N.}
$$
Its single-particle <spin> and <parity> is $1/2^-$. The negative moment results from the proton spin being predominantly opposed to the total <angular momentum> in this $j=\ell-1/2$ level.
For oxygen, eight <protons> and the first eight <neutrons> are paired. The ninth <neutron> occupies $1d_{5/2}$, with $\ell=2$, $j=5/2=\ell+1/2$. Since its orbital factor is zero, $g_j=g_n/(2\ell+1)=-3.826/5$. Therefore
$$
\boxed{\mu({}^{17}\mathrm O)=-1.913\,\mu_N,\qquad J^P=5/2^+.}
$$
These are ideal single-particle predictions in units of the <nuclear magneton>, not fits incorporating nuclear configuration mixing.
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