Solution:

Let the points be $0, \ldots, 7 \pmod{8}$, and view Alice's reveal as revealing the three possible locations of the apple. If Alice always picks $0,2,4$ and puts the apple randomly at $0$ or $4$, by symmetry Bob cannot achieve more than $\frac{1}{2}$. Here's a proof that $\frac{1}{2}$ is always possible.

Among the three revealed indices $a, b, c$, positioned on a circle, two must (in the direction in which they're adjacent) have distance at least $3$, so without loss of generality the three are $0, b, c$ where $1 \leq b < c \leq 5$. Modulo reflection and rotation, the cases are:

$(0,1,2)$: Bob places at $1$ and wins.

$(0,1,3)$: Bob places at $1$ half the time and $3$ half the time, so wherever the apple is Bob wins with probability $\frac{1}{2}$.

$(0,1,4)$: Bob places at $1$ or $4$, same as above.

$(0,2,4)$: Bob places at $1$ or $3$, same as above.

$(0,2,5)$: Bob places at $1$ or $5$, same as above.

These cover all cases, so we're done.