The given function represents a discrete uniform distribution with mean 1. The sample mean of a random sample of size 36 taken with replacement from this population will also have a mean of 1 and a standard deviation of $\frac{1}{\sqrt{36}}=\frac{1}{6}$.

To find the probability that the sample mean is greater than 1.4 but less than 1.8, we need to find the probability that the sample mean falls within the range of 1.4 to 1.8. We can standardize this range using the formula:

$z=\frac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\frac{\bar{x}-1}{1/6}$

where $\bar{x}$ is the sample mean, $\mu$ is the population mean (which is 1), $\sigma$ is the population standard deviation (which is $\frac{1}{6}$), and $n$ is the sample size (which is 36).

Substituting the values, we get:

$z=\frac{\bar{x}-1}{1/6}=\frac{1.4-1}{1/6}=0.4\times6=2.4$

and

$z=\frac{\bar{x}-1}{1/6}=\frac{1.8-1}{1/6}=0.8\times6=4.8$

Using a standard normal distribution table or calculator, we can find the probability that $z$ falls between 2.4 and 4.8. This probability is approximately 0.0136.

Therefore, the probability that for a random sample of 36 taken with replacement from the given population, the sample mean is greater than 1.4 but less than 1.8 is approximately 0.0136.