The heights of people h metres in a community are normally distributed with mean µ and standard deviation σ. It is given that P(h < 1.2) = 0.15 and P(h > 1.6) = 0.10. Calculate the mean µ and standard deviation σ of the heights of the people.

Given $P(h < 1.2) = 0.15$. We standardize this to $P\left(Z < \frac{1.2 - \mu}{\sigma}\right) = 0.15$.
From the standard normal distribution table (or calculator), the z-score corresponding to a cumulative probability of 0.15 is $z_1 \approx -1.0364$.
So, we have the equation:

Given $P(h > 1.6) = 0.10$. This means $P(h < 1.6) = 1 - 0.10 = 0.90$. We standardize this to $P\left(Z < \frac{1.6 - \mu}{\sigma}\right) = 0.90$.
From the standard normal distribution table, the z-score corresponding to a cumulative probability of 0.90 is $z_2 \approx 1.2816$.
So, we have the equation:

From (1): $1.2 - \mu = -1.0364\sigma$
From (2): $1.6 - \mu = 1.2816\sigma$
Subtract equation (1) from equation (2):

Substitute the value of $\sigma$ back into the first equation:

Rounding to three decimal places:
The mean $\mu$ is $\boxed{\text{1.379 m}}$
The standard deviation $\sigma$ is $\boxed{\text{0.173 m}}$

The lower quartile ($Q_1$) corresponds to $P(h < Q_1) = 0.25$. The z-score for 0.25 is $z_{Q_1} \approx -0.6745$.
The upper quartile ($Q_3$) corresponds to $P(h < Q_3) = 0.75$. The z-score for 0.75 is $z_{Q_3} \approx 0.6745$.

Rounding to three decimal places:
The lower quartile is $\boxed{\text{1.262 m}}$
The upper quartile is $\boxed{\text{1.495 m}}$

The inequality $|h - \mu| < 0.1$ means $-0.1 < h - \mu < 0.1$.
Divide by $\sigma$:

Using $\sigma \approx 0.17256$:

Since the normal distribution is symmetric, $P(Z < -z) = 1 - P(Z < z)$.

From the standard normal distribution table, $P(Z < 0.5795) \approx 0.7189$.

Rounding to three decimal places:
$P(|h - \mu| < 0.1)$ is $\boxed{\text{0.438}}$