Let the probability of rating the service as 'poor' or 'satisfactory' be the sum of the probabilities of these two ratings:

\(p = 0.12 + 0.36 = 0.48\).

We need to find the probability that more than 2 and fewer than 8 residents rate the service as 'poor' or 'satisfactory'. This is equivalent to finding \(P(3 \leq X \leq 7)\) where \(X\) is a binomial random variable with parameters \(n = 8\) and \(p = 0.48\).

Using the complement rule, we have:

\(P(3 \leq X \leq 7) = 1 - P(X = 0, 1, 2, 8)\)

Calculate \(P(X = 0, 1, 2, 8)\):

\(P(X = 0) = \binom{8}{0} (0.48)^0 (0.52)^8 = 0.00534597\)

\(P(X = 1) = \binom{8}{1} (0.48)^1 (0.52)^7 = 0.039478\)

\(P(X = 2) = \binom{8}{2} (0.48)^2 (0.52)^6 = 0.127544\)

\(P(X = 8) = \binom{8}{8} (0.48)^8 (0.52)^0 = 0.0028179\)

Thus,

\(P(X = 0, 1, 2, 8) = 0.00534597 + 0.039478 + 0.127544 + 0.0028179 = 0.1752\)

Therefore,

\(P(3 \leq X \leq 7) = 1 - 0.1752 = 0.8248\)

Thus, the probability that more than 2 and fewer than 8 residents rate the service as 'poor' or 'satisfactory' is approximately 0.825.