The question is asking for the probability that a random variable $X$, which follows a normal distribution with a mean $\mu$ of 40 and a standard deviation $\sigma$ of 14, does not lie between the values 12 and 61. To find this probability, we first calculate the standardized $Z$-scores for the given values using the formula:

$Z = \frac{X - \mu}{\sigma}$

For $X = 12$, the $Z$-score is:

$Z = \frac{12 - 40}{14} = -2$

For $X = 61$, the $Z$-score is:

$Z = \frac{61 - 40}{14} = 1.5$

Using a standard normal distribution table (Z-table), we find the probabilities for $Z$ being less than -2 and greater than 1.5. The probability $P(Z < -2)$ is approximately 0.0228, and $P(Z > 1.5)$ is approximately 0.0668. Since we are looking for the probability that $X$ does not lie between 12 and 61, we are interested in the combined probability of $Z$ being less than -2 or greater than 1.5.

The combined probability is calculated by adding the individual probabilities:

$P(Z < -2) + P(Z > 1.5) = 0.0228 + 0.0668 = 0.0896$

This gives us the probability that $X$ does not lie between 12 and 61 as 8.96%, which corresponds to option C. This question tests the understanding of how to work with probabilities in the context of a normal distribution and the use of Z-tables to find probabilities associated with specific $Z$-scores.