= Solution
Events $A$ and $B$ are <independent events>[independent] when
$$
\mathbb P(A\cap B)=\mathbb P(A)\mathbb P(B).
$$
For a possible sum $2\leq t\leq12$, let $N_t$ be the number of ordered die pairs with sum $t$. Then
$$
\mathbb P(A_t)=\frac{N_t}{36},
\qquad
\mathbb P(B_i)=\frac16.
$$
If $t-i\in\{1,\ldots,6\}$, then $\mathbb P(A_t\cap B_i)=1/36$; otherwise it is zero. Independence for a possible sum therefore requires
$$
\frac1{36}=\frac{N_t}{216},
$$
so $N_t=6$. This occurs only for $t=7$, and then $7-i$ is valid for every $i=1,\ldots,6$. Thus
$$
\boxed{A_t\text{ and }B_i\text{ are independent exactly when }
t=7,\ 1\leq i\leq6}.
$$
If impossible sums are admitted as events of probability zero, they are trivially independent as well.
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