Step 1: Fix the oxidation number of oxygen.
In $Br_3O_8$ each oxygen carries $-2$, and there are $8$ oxygens, giving a total of $8 \times (-2) = -16$.
Step 2: Balance the neutral molecule.
The molecule has no overall charge, so the three bromine oxidation states must add to $+16$.
Step 3: See why the sum alone is not enough.
Several option sets ($5+6+5$, $6+4+6$, $7+2+7$, $6+3+7$) all sum to $16$, so we must use the actual structure to decide.
Step 4: Use the structure of $Br_3O_8$.
The molecule has a central bromine bridging two terminal bromine atoms through oxygen, with the two terminal bromines in a higher state and the central one in a lower state.
Step 5: Assign the values.
The two terminal bromines are $+6$ each and the central bromine is $+4$, giving $6 + 4 + 6 = 16$, consistent with the neutral molecule.
Step 6: State the answer.
The oxidation states are $+6, +4, +6$, which is option (2). \[ \boxed{+6,\,+4,\,+6} \]