= Solution
For three players, the predecessor-set weights are $1/3$ for sizes zero and two, and $1/6$ for size one. Thus
$$
\begin{aligned}
\phi_1&=\tfrac13(4)+\tfrac16(10-3)+\tfrac16(11-2)+\tfrac13(12-5)=\tfrac{19}3,\\
\phi_2&=\tfrac13(3)+\tfrac16(10-4)+\tfrac16(5-2)+\tfrac13(12-11)=\tfrac{17}6,\\
\phi_3&=\tfrac13(2)+\tfrac16(11-4)+\tfrac16(5-3)+\tfrac13(12-10)=\tfrac{17}6.
\end{aligned}
$$
Therefore \b[the <Shapley value> is $\boxed{(19/3,17/6,17/6)}$], whose coordinates sum to $12$.
For the <nucleolus>, use <imputations> $x_1+x_2+x_3=12$, $x_1\geq4$, $x_2\geq3$, $x_3\geq2$. The excess of $\{1,3\}$ is $11-(12-x_2)=x_2-1\geq2$. Thus the smallest possible largest excess is at least $2$. It is attained when $x_2=3$, $x_3=9-x_1$, and $5\leq x_1\leq7$: the remaining proper-coalition excesses then are
$$
e_1=4-x_1,\quad e_2=0,\quad e_3=x_1-7,\quad e_{12}=7-x_1,\quad e_{23}=x_1-7,
$$
all at most $2$, while $e_{13}=2$. Conversely, a largest excess of $2$ forces $x_2=3$ and precisely this interval for $x_1$.
On this first-stage face the top excess $2$ is fixed. The next largest excess is $7-x_1\geq0$, because the other varying excesses are nonpositive and $e_2=0$ is fixed. Its unique minimum is zero at $x_1=7$. No further lexicographic tie remains. Hence \b[the <nucleolus> is $\boxed{(7,3,2)}$]. Its proper-coalition excesses, sorted decreasingly, are $(2,0,0,0,0,-3)$.
Finally, a core allocation would require $x_2\geq3$ and $x_1+x_3\geq11$, whose sum contradicts the efficient total $12$. Thus \b[the <core of a cooperative game> is empty]. The game is not superadditive, so neither core nonemptiness nor individual rationality of the <Shapley value> should be presumed. Under the distinct <prenucleolus> convention the answer would be $(15/2,2,5/2)$. Balancing $e_2=3-x_2$ and $e_{13}=x_2-1$ gives a first-stage maximum of $1$ and forces $x_2=2$. The remaining first-stage constraints restrict $x_1$ to $[7,8]$. The next largest complaints are $e_{12}=8-x_1$ and $e_{23}=x_1-7$, balanced at $x_1=15/2$, with $x_3=5/2$. This is not an <imputation> and is not the <nucleolus> under the definition in (a).
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