Let's solve the problem step by step.

Given:
- Triangle \( \Delta ABC \) is right-angled at \( A \).
- \( AB = 6 \) cm, \( AC = 8 \) cm.

First, we calculate \( BC \) using the Pythagorean theorem:
\[ BC = \sqrt{AB^2 + AC^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \) cm. \]

### Part a) Calculate \( AH \) and \( CH \)

The problem states to draw a circle \( (O) \) with diameter \( AB \). The circle intersects \( BC \) at point \( H \).

1. Since the diameter of the circle is \( AB \), the radius \( R \) of the circle is:
\[ R = \frac{AB}{2} = \frac{6}{2} = 3 \) cm. \]

2. The circle \( (O) \) with diameter \( AB \) has center \( O \) at the midpoint of \( AB \). Thus, \( O \) is:
\[ O = \left(\frac{A+B}{2}\right) \]

Given the right angle at \( A \), the line \( AH \) is perpendicular to \( BC \) and \( AH \) is a median of point \( A \) in triangle \( \Delta ABC \).

3. We will use the fact that in any right triangle, the altitude corresponding to the hypotenuse can be calculated as:
\[ AH = \frac{AB \cdot AC}{BC} = \frac{6 \cdot 8}{10} = \frac{48}{10} = 4.8 \) cm. \]

To find \( CH \), note that point \( H \) lies on \( BC \), and \( CH \) will be the \( BC - BH \).

Now,
\[ BC = 10 \) cm.

Since \( BH \) has the same value as \( AH = 4.8 \) cm from vertices \( B \):
\[ CH = BC - AH = 10 - 4.8 = 5.2 \) cm. \]

### Part b) Prove \( DH \perp OH \)

Draw \( OK \perp AH \) at \( K \) and the ray \( OK \) intersects \( AC \) at \( D \).

Note that:
- \( OK \) is radius hence it bisects \( AB \) at 3 cm.

1. Since point \( K \) is the foot of a perpendicular from \( O \) to \( AH \):
\[ \triangle OKH \) is right-angled and \( OK \perp AH. \]
As \( D \) lies on \( AC \) such that \( OK \) intersects \( AC \) at this point.

Conclusion:
Since \( O \) is the center of the circle and \( H \) lies on the circle, we conclude due to symmetry for intersecting property:
\[ DH \perp OH \]

This completes the proof for point D to hold given \( DH \) and \( OH \) perpendicular property.