Calculate the value of x and the size of alpha.

The distance $OP$ is given as 2.

$$OP = \sqrt{(x_P - x_O)^2 + (y_P - y_O)^2}$$ $$2 = \sqrt{(x - 0)^2 + (\sqrt{3} - 0)^2}$$ $$2 = \sqrt{x^2 + (\sqrt{3})^2}$$

$$2 = \sqrt{x^2 + 3}$$

Square both sides:

$$2^2 = x^2 + 3$$ $$4 = x^2 + 3$$ $$x^2 = 4 - 3$$ $$x^2 = 1$$ $$x = \pm 1$$

Therefore, $x = 1$.
The value of $x$ is $\boxed{1}$.

1.1.2 Hence, calculate the size of $\alpha$.

Let this angle be $\theta_P$.
We know $OP = 2$.

$$\cos \theta_P = \frac{x}{OP} = \frac{1}{2}$$ $$\sin \theta_P = \frac{y}{OP} = \frac{\sqrt{3}}{2}$$

This means $\theta_P = 60^\circ$.

From the diagram, point $Q$ is in the fourth quadrant, which means the angle for $Q$ is obtained by rotating $P$ clockwise by $90^\circ$.
Let $\theta_Q$ be the angle for point $Q$.

$$\theta_Q = \theta_P - 90^\circ$$ $$\theta_Q = 60^\circ - 90^\circ$$ $$\theta_Q = -30^\circ$$

Since $\theta_Q = -30^\circ$, this means $OQ$ is $30^\circ$ clockwise from the positive $x$-axis.
Therefore, $\alpha = 30^\circ$.
The size of $\alpha$ is $\boxed{30^\circ}$.

1.1.3 Determine the coordinates of $Q$.

We know $OQ = 20$ and the angle for $Q$ is $\theta_Q = -30^\circ$.
The coordinates $(a; b)$ can be found using:

$$a = OQ \cos \theta_Q$$ $$b = OQ \sin \theta_Q$$

$$a = 20 \cos(-30^\circ)$$ $$b = 20 \sin(-30^\circ)$$

$$\cos(-30^\circ) = \cos(30^\circ) = \frac{\sqrt{3}}{2}$$ $$\sin(-30^\circ) = -\sin(30^\circ) = -\frac{1}{2}$$

$$a = 20 \left(\frac{\sqrt{3}}{2}\right) = 10\sqrt{3}$$ $$b = 20 \left(-\frac{1}{2}\right) = -10$$

The coordinates of $Q$ are $\boxed{(10\sqrt{3}; -10)}$.