Find CBD, ACD, and AXB given the diagram with angles CAD = 33^ and ACB = 42^, and BD is a diameter.

Angles $\angle CAD$ and $\angle CBD$ are subtended by the same arc CD.

Angles subtended by the same arc at the circumference are equal.

Given $\angle CAD = 33^\circ$.

The value of $\angle CBD$ is $\boxed{\text{33}^\circ}$.

Since BD is the diameter, the angle subtended by the diameter at any point on the circumference is $90^\circ$.
Therefore, $\angle BCD = 90^\circ$.

The angle $\angle BCD$ can be expressed as the sum of $\angle ACB$ and $\angle ACD$.

Given $\angle ACB = 42^\circ$.

The value of $\angle ACD$ is $\boxed{\text{48}^\circ}$.

We need to find two angles in triangle AXD to determine the third.
$\angle XAD = \angle CAD = 33^\circ$ (given).

$\angle XDA$ is the same as $\angle BDA$.
Angles $\angle BDA$ and $\angle BCA$ are subtended by the same arc AB.
Therefore, $\angle BDA = \angle BCA$.
Given $\angle BCA = 42^\circ$.
So, $\angle XDA = 42^\circ$.

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

Angles $\angle AXB$ and $\angle AXD$ form a linear pair on the straight line BD, so they are supplementary.

The value of $\angle AXB$ is $\boxed{\text{75}^\circ}$.