In the diagram, A,B,C and D are points on a circle centre O. BD is a diameter and AF is the tangent to the circle at A. CBD = 30° and AC and BD intersect at E such that AEB = 70°. Find CAD, ADB, ACD, CAF.

Angles subtended by the same arc in a circle are equal.

$$\angle CAD = \angle CBD$$

$$\angle CAD = 30^\circ$$

The value of $\angle CAD$ is $\boxed{30^\circ}$.

Angles $\angle AEB$ and $\angle AED$ are supplementary because they form a linear pair on the straight line BD.

$$\angle AED = 180^\circ - \angle AEB$$ $$\angle AED = 180^\circ - 70^\circ = 110^\circ$$

In $\triangle ADE$, the sum of angles is $180^\circ$. We know $\angle DAE = \angle CAD = 30^\circ$ from part (i).

$$\angle DAE + \angle ADE + \angle AED = 180^\circ$$ $$30^\circ + \angle ADB + 110^\circ = 180^\circ$$

$$\angle ADB = 180^\circ - 30^\circ - 110^\circ$$ $$\angle ADB = 40^\circ$$

The value of $\angle ADB$ is $\boxed{40^\circ}$.

Since BD is a diameter, the angle subtended by the diameter at any point on the circumference is $90^\circ$.
Therefore, $\angle DAB = 90^\circ$.
In $\triangle DAB$, the sum of angles is $180^\circ$. We know $\angle ADB = 40^\circ$ from part (ii).

$$\angle DAB + \angle ADB + \angle ABD = 180^\circ$$ $$90^\circ + 40^\circ + \angle ABD = 180^\circ$$

$$\angle ABD = 180^\circ - 90^\circ - 40^\circ$$ $$\angle ABD = 50^\circ$$

Angles subtended by the same arc in a circle are equal.

$$\angle ACD = \angle ABD$$ $$\angle ACD = 50^\circ$$

The value of $\angle ACD$ is $\boxed{50^\circ}$.

AF is a tangent to the circle at A, and AC is a chord. The angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment.

$$\angle CAF = \angle ABC$$

$\angle ABC$ is the sum of $\angle ABD$ and $\angle DBC$.
We found $\angle ABD = 50^\circ$ in part (iii), and $\angle DBC = 30^\circ$ is given.

$$\angle ABC = \angle ABD + \angle DBC$$ $$\angle ABC = 50^\circ + 30^\circ$$ $$\angle ABC = 80^\circ$$

$$\angle CAF = 80^\circ$$

The value of $\angle CAF$ is $\boxed{80^\circ}$.