The context for question 13, including the diagram or description of the setup related to lines XY and OC, is missing from the provided image. Therefore, it is not possible to answer parts (i), (ii),

The angle given in the diagram ($50^\circ$) is between the incident ray EO and the surface of the block. The angle of incidence ($\theta_1$) is measured from the normal to the surface.

$$\theta_1 = 90^\circ - 50^\circ = 40^\circ$$

The angle of refraction ($\theta_2$) is given as $18^\circ$.
The refractive index of air ($n_1$) is approximately 1.

Here, $n_1 = 1$ (for air), $\theta_1 = 40^\circ$, $n_2$ is the refractive index of the block, and $\theta_2 = 18^\circ$.

$$1 \times \sin(40^\circ) = n_{block} \times \sin(18^\circ)$$

$$n_{block} = \frac{\sin(40^\circ)}{\sin(18^\circ)}$$ $$n_{block} = \frac{0.6428}{0.3090}$$ $$n_{block} \approx 2.080$$

The refractive index of the block is $\boxed{\text{2.08}}$.