= Solution
Closed subshells have $L=S=0$, so only the open $p$ or $d$ subshell determines each <atomic term>. Apply <Hund's rules> in order: maximize $S$, maximize $L$ at that $S$, then choose the lowest $J$ below half filling or the highest $J$ above half filling.
For <carbon>, the two $2p$ <electrons> occupy different magnetic <atomic orbitals> with parallel <spins>, giving $S=1$. Their largest allowed $M_L$ at that <spin> is $1+0=1$, so $L=1$, a $P$ term. The $p^2$ subshell is less than half full, giving $J=|1-1|=0$. \b[The ground level is ${}^3P_0$.]
For <vanadium>, three parallel-spin $3d$ <electrons> give $S=3/2$. The largest $M_L$ compatible with the <Pauli exclusion principle> is $2+1+0=3$, hence $L=3$, an $F$ term. The $d^3$ subshell is less than half full, so $J=3-3/2=3/2$. \b[The ground level is ${}^4F_{3/2}$.]
For <cobalt>, the $d^7$ subshell is equivalent to three $d$ holes for the allowed $L,S$ terms, so its maximum-spin, maximum-$L$ term is again ${}^4F$. Explicitly, five spin-up <electrons> fill all $m_l$ values and two spin-down <electrons> can occupy $m_l=2,1$, giving $M_S=3/2$ and $M_L=3$. Since $d^7$ is more than half full, $J=3+3/2=9/2$. \b[The ground level is ${}^4F_{9/2}$.]
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