The distance X travelled by a bus on a full tank of diesel has a probability density function given by f(x) = { kx 0 ≤ x ≤ 1, k 1 < x ≤ 3, 0 otherwise, where k is a constant. (i) Find the value of k. (ii) Show that the median m of x is 1.75. Hence find P(|X - m| < 1/2).

For a probability density function, the integral of $f(x)$ over its entire domain must equal 1.

$$\int_{-\infty}^{\infty} f(x) \,dx = 1$$

Given the function:

$$f(x)=\begin{cases} kx & 0 \le x \le 1 \\ k & 1 < x \le 3 \\ 0 & \text{otherwise} \end{cases}$$

We integrate $f(x)$ over its defined intervals:

$$\int_{0}^{1} kx \,dx + \int_{1}^{3} k \,dx = 1$$

First integral:

$$\int_{0}^{1} kx \,dx = k \left[ \frac{x^2}{2} \right]_{0}^{1} = k \left( \frac{1^2}{2} - \frac{0^2}{2} \right) = k \left( \frac{1}{2} \right) = \frac{k}{2}$$

Second integral:

$$\int_{1}^{3} k \,dx = k \left[ x \right]_{1}^{3} = k (3 - 1) = 2k$$

$$\frac{k}{2} + 2k = 1$$ $$\frac{k}{2} + \frac{4k}{2} = 1$$ $$\frac{5k}{2} = 1$$ $$5k = 2$$ $$k = \frac{2}{5}$$

The value of $k$ is $\boxed{\frac{2}{5}}$.

The median $m$ is the value such that $P(X \le m) = 0.5$.
Using $k = \frac{2}{5}$, the probability density function is:

$$f(x)=\begin{cases} \frac{2}{5}x & 0 \le x \le 1 \\ \frac{2}{5} & 1 < x \le 3 \\ 0 & \text{otherwise} \end{cases}$$

First, check the probability up to $x=1$:

$$P(X \le 1) = \int_{0}^{1} \frac{2}{5}x \,dx = \frac{2}{5} \left[ \frac{x^2}{2} \right]_{0}^{1} = \frac{2}{5} \left( \frac{1}{2} \right) = \frac{1}{5} = 0.2$$

Since $P(X \le 1) = 0.2 < 0.5$, the median $m$ must be in the interval $1 < x \le 3$.
So, we set up the integral for $P(X \le m) = 0.5$:

$$P(X \le m) = \int_{0}^{1} \frac{2}{5}x \,dx + \int_{1}^{m} \frac{2}{5} \,dx = 0.5$$ $$0.2 + \int_{1}^{m} \frac{2}{5} \,dx = 0.5$$ $$\int_{1}^{m} \frac{2}{5} \,dx = 0.3$$

$$\frac{2}{5} \left[ x \right]_{1}^{m} = 0.3$$ $$\frac{2}{5} (m - 1) = 0.3$$ $$m - 1 = 0.3 \times \frac{5}{2}$$ $$m - 1 = \frac{3}{10} \times \frac{5}{2}$$ $$m - 1 = \frac{15}{20}$$ $$m - 1 = \frac{3}{4}$$ $$m - 1 = 0.75$$ $$m = 1 + 0.75$$ $$m = 1.75$$

Thus, the median $m$ is indeed $\boxed{1.75}$.

The inequality $|X-m|<\frac{1}{2}$ can be rewritten as:

$$-\frac{1}{2} < X - m < \frac{1}{2}$$ $$m - \frac{1}{2} < X < m + \frac{1}{2}$$

Substitute $m = 1.75$:

$$1.75 - 0.5 < X < 1.75 + 0.5$$ $$1.25 < X < 2.25$$

Both 1.25 and 2.25 fall within the interval $1 < x \le 3$, where $f(x) = \frac{2}{5}$.

$$P(1.25 < X < 2.25) = \int_{1.25}^{2.25} \frac{2}{5} \,dx$$ $$P(1.25 < X < 2.25) = \frac{2}{5} \left[ x \right]_{1.25}^{2.25}$$ $$P(1.25 < X < 2.25) = \frac{2}{5} (2.25 - 1.25)$$ $$P(1.25 < X < 2.25) = \frac{2}{5} (1)$$ $$P(1.25 < X < 2.25) = \frac{2}{5}$$

The probability is $\boxed{\frac{2}{5}}$.