X is a continuous random variable with probability density function f(x). The f(x) is defined by a curve y = x²; 0 ≤ x ≤1 and a straight line as indicated on the diagram. (i) Find the value of h. (ii) Write down the mode. (iii) Find the lower quartile.

For $0 \le x \le 1$, $f(x) = x^2$.
For $1 \le x \le h$, the function is a straight line connecting $(1,1)$ and $(h,0)$.
The equation of this line is $f(x) = \frac{0-1}{h-1}(x-h) = \frac{-(x-h)}{h-1} = \frac{h-x}{h-1}$.

This integral can be split into two parts:

The first integral:

The second integral represents the area of a triangle with base $(h-1)$ and height 1:

Multiply by 6 to clear the denominators:

The value of $h$ is $\boxed{\frac{7}{3}}$.

The lower quartile $Q_1$ is the value such that $P(X \le Q_1) = 0.25$.
This means $\int_{0}^{Q_1} f(x)\,dx = 0.25$.

We know that $\int_{0}^{1} f(x)\,dx = \frac{1}{3} \approx 0.333$.
Since $0.25 < \frac{1}{3}$, the lower quartile $Q_1$ must lie in the interval $0 \le x \le 1$.
In this interval, $f(x) = x^2$.

Rounding to 3 significant figures, the lower quartile is $\boxed{0.909}$.