In the diagram, the points A, B, C and D lie on the circumference of a circle, centre O. The line ST is the tangent at A and the radius AO, when produced, meets DC at P. Given that angle ADC = 75° and angle APD = 50°.

ABCD is a cyclic quadrilateral. The opposite angles of a cyclic quadrilateral sum to $180^\circ$.

$$\angle ABC + \angle ADC = 180^\circ$$

Given $\angle ADC = 75^\circ$.

$$\angle ABC = 180^\circ - 75^\circ$$ $$\angle ABC = \boxed{\text{105}^\circ}$$

In $\triangle APD$:
The sum of angles in a triangle is $180^\circ$.

$$\angle PAD + \angle APD + \angle ADP = 180^\circ$$

Given $\angle APD = 50^\circ$.
Since D, P, C are collinear, $\angle ADP = \angle ADC = 75^\circ$.

$$\angle PAD + 50^\circ + 75^\circ = 180^\circ$$ $$\angle PAD = 180^\circ - 125^\circ$$ $$\angle PAD = 55^\circ$$

The radius AO is perpendicular to the tangent ST at point A.

$$\angle SAO = 90^\circ$$ $$\angle SAD = \angle SAO - \angle DAO$$

Since P lies on the extension of AO, $\angle DAO = \angle PAD = 55^\circ$.

$$\angle SAD = 90^\circ - 55^\circ$$ $$\angle SAD = \boxed{\text{35}^\circ}$$

In $\triangle OAD$, OA and OD are radii, so $\triangle OAD$ is an isosceles triangle.

$$\angle ODA = \angle OAD = 55^\circ$$

Now, find $\angle ODC$:

$$\angle ODC = \angle ADC - \angle ODA$$ $$\angle ODC = 75^\circ - 55^\circ$$ $$\angle ODC = 20^\circ$$

In $\triangle ODC$, OC and OD are radii, so $\triangle ODC$ is an isosceles triangle.

$$\angle OCD = \angle ODC = 20^\circ$$

In $\triangle OPC$:
The sum of angles in a triangle is $180^\circ$.

$$\angle POC + \angle OPC + \angle OCP = 180^\circ$$

$\angle OPC$ is the same as $\angle APD = 50^\circ$.
$\angle OCP$ is the same as $\angle OCD = 20^\circ$.

$$\angle POC + 50^\circ + 20^\circ = 180^\circ$$ $$\angle POC = 180^\circ - 70^\circ$$ $$\angle POC = \boxed{\text{110}^\circ}$$

The diagonals of a rhombus intersect at right angles. Therefore, $\angle EKH = 90^\circ$.
In $\triangle EHK$, the sum of angles is $180^\circ$.

$$\angle HEK + \angle EHK + \angle EKH = 180^\circ$$

Given $\angle HEK = x^\circ$.

$$x^\circ + \angle EHK + 90^\circ = 180^\circ$$ $$\angle EHK = 180^\circ - 90^\circ - x^\circ$$ $$\angle EHK = \boxed{(90 - x)^\circ}$$

The diagonals of a rhombus bisect the angles of the rhombus.
Given $\angle HEK = x^\circ$. This means $\angle HEK$ is half of $\angle FEH$.

$$\angle FEH = 2 \times \angle HEK$$ $$\angle FEH = 2x^\circ$$

Opposite angles of a rhombus are equal.

$$\angle FGH = \angle FEH$$ $$\angle FGH = \boxed{2x^\circ}$$