Step 1: Fix the condition.
We count every $8$ that is divisible by both its left neighbour and its right neighbour.
Step 2: Recall what divides 8.
Among small numbers, $8$ is divisible by $1,2,4$ but not by $3$ or $6$.
Step 3: Scan early triplets.
$2,8,3$ fails ($3\nmid 8$); $3,8,2$ fails; the triplet $4,8,2$ works since $4\mid 8$ and $2\mid 8$. That is one.
Step 4: Continue scanning.
$4,8,6$ fails ($6\nmid 8$); $6,8,2$ fails; the triplet $2,8,2$ works. That is two.
Step 5: Finish the scan.
$4,8,3$ fails, $3,8,2$ fails, $2,8,6$ fails, so no further valid $8$ appears.
Step 6: Total the count.
Exactly two $8$s satisfy the rule.
\[ \boxed{2} \]