= Solution
First restrict to <integers> in the marginal supports: $2\leq i\leq12$ and $-5\leq j\leq5$. Their marginal <probabilities> are positive. By part (i), a feasible joint pair identifies a unique ordered pair of rolls, so its joint <probability> is $1/36$; an infeasible pair has joint <probability> zero.
Put $a_i=6-|i-7|$ and $b_j=6-|j|$. If the two <events> were independent, a feasible pair would require
$$
\frac1{36}=\frac{a_i}{36}\frac{b_j}{36},\qquad\text{so }a_ib_j=36.
$$
Since $1\leq a_i,b_j\leq6$, equality forces $a_i=b_j=6$, hence $(i,j)=(7,0)$. But that pair is infeasible. An infeasible pair cannot give independence either, since its joint <probability> is zero while the product of its positive marginal <probabilities> is positive. Thus
$$
\boxed{\text{There are no independent pairs of these events within the two marginal supports}.}
$$
If all real values of $i,j$ are allowed literally, an impossible <event> is independent of every <event>. The complete answer without a support restriction is therefore
$$
\boxed{\{X=i\}\text{ and }\{Y=j\}\text{ are independent iff }i\notin\{2,\ldots,12\}\text{ or }j\notin\{-5,\ldots,5\}.}
$$
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