To solve the problems given in the triangle ABC, we can proceed as follows:

### Problem 68

1. **Given:**
- \( AB = 3 \, \text{cm} \)
- \( BC = 5 \, \text{cm} \)
- Right triangle at \( A \) (so \( AC \) is the hypotenuse).

2. **Part a: Calculate \( HM \) and \( BM \)**:
- Use the Pythagorean theorem to find \( AC \):
\[
AC = \sqrt{AB^2 + BC^2} = \sqrt{3^2 + 5^2} = \sqrt{9 + 25} = \sqrt{34}.
\]

3. **Part b: Prove that \( AH \cdot BC = HN \cdot AC + HM \cdot AB \)**:
- We can apply the property of right triangles or similar triangles. \( AH \) can be calculated as:
\[
AH = \frac{AB \cdot BC}{AC}.
\]

4. **Part c: Identify \( Q \) and \( K \)**:
- \( Q \) is the midpoint of \( BH \) and \( K \) is the midpoint of \( CH \).
- To prove \( MQ \parallel KN \), show that the triangles \( BHM \) and \( CHN \) are similar due to the right angles and shared angles.

### Problem 69

1. **Given:**
- \( AH \) and \( BK \) are heights from points \( A \) and \( B \), respectively.

2. **Part a:**
- To show that \( BD = 2AH \) when \( D \) is the foot of the altitude from \( B \) to \( AC \):
\[
BD = \text{the height from } B \text{ to line } AC \text{ can be shown to equal } 2AH \text{ through similarity or area arguments.}
\]

3. **Part b: Prove the relation:**
\[
\frac{1}{BK^2} = \frac{1}{BC^2} + \frac{1}{4AH^2}
\]
- You can employ the properties of similar triangles and their corresponding heights to establish this relationship.

### General Advice

While you work through these problems, drawing diagrams and labeling all relevant points and lengths can help visualize the relationships and make the required proofs easier to follow.