= Solution
The 36 ordered outcomes are equiprobable. For $2\le s\le12$, let $c_s=6-|s-7|$ count outcomes with sum $s$. Then $\mathbb P(A_s)=c_s/36$ and $\mathbb P(B_i)=1/6$ for $1\le i\le6$. Their joint <probability> is $1/36$ if $1\le s-i\le6$, and zero otherwise.
If the joint <probability> is zero, independence is impossible in these ranges because both marginal probabilities are positive. If it is $1/36$, the <independent events> condition requires $1/36=c_s/216$, or $c_s=6$. This happens only for $s=7$, and then all six values of $i$ allow the partner $7-i$. Therefore \b[within the possible sums and faces]
$$
\boxed{s=7,\qquad i\in\{1,2,3,4,5,6\}.}
$$
If arbitrary integer labels outside these ranges are included, an impossible sum or face defines an empty <event>, which is trivially independent of every <event>.
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