Let’s solve the problem step by step based on the image you uploaded.

a) Express $\overrightarrow{MN}$ in terms of $a$ and $b$:

1. Given Information:

   • $\overrightarrow{AM} = 5a$
   • $\overrightarrow{AC} = 9b$
   • $M$ is the midpoint of $AB$, and $N$ is the midpoint of $BC$.

2. Position vectors: Let $\overrightarrow{A}$, $\overrightarrow{B}$, and $\overrightarrow{C}$ represent the position vectors of the vertices $A$, $B$, and $C$, respectively. Then:

   • $\overrightarrow{M} = \frac{\overrightarrow{A} + \overrightarrow{B}}{2}$ (midpoint of $AB$).
   • $\overrightarrow{N} = \frac{\overrightarrow{B} + \overrightarrow{C}}{2}$ (midpoint of $BC$).

3. Expressing $\overrightarrow{B}$ in terms of $a$ and $b$: Using $\overrightarrow{AM} = 5a$: $\overrightarrow{B} = \overrightarrow{M} - \overrightarrow{A} = 5a + \overrightarrow{A}.$ Using $\overrightarrow{AC} = 9b$: $\overrightarrow{C} = \overrightarrow{A} + 9b.$

4. Find $\overrightarrow{MN}$: Substitute $\overrightarrow{M}$ and $\overrightarrow{N}$: $\overrightarrow{M} = \frac{\overrightarrow{A} + (5a + \overrightarrow{A})}{2} = \frac{2\overrightarrow{A} + 5a}{2},$ $\overrightarrow{N} = \frac{(5a + \overrightarrow{A}) + (\overrightarrow{A} + 9b)}{2} = \frac{2\overrightarrow{A} + 5a + 9b}{2}.$ Then, $\overrightarrow{MN} = \overrightarrow{N} - \overrightarrow{M} = \frac{2\overrightarrow{A} + 5a + 9b}{2} - \frac{2\overrightarrow{A} + 5a}{2}.$ Simplify: $\overrightarrow{MN} = \frac{9b}{2}.$

Final Answer: $\overrightarrow{MN} = \frac{9b}{2}.$

b) Prove $MN \parallel AC$ and calculate the ratio $\frac{AC}{MN}$:

1. Direction of $\overrightarrow{MN}$ and $\overrightarrow{AC}$:

   • From part (a), $\overrightarrow{MN} = \frac{9b}{2}$.
   • From the problem, $\overrightarrow{AC} = 9b$.

   Clearly, $\overrightarrow{MN}$ is a scalar multiple of $\overrightarrow{AC}$: $\overrightarrow{MN} = \frac{1}{2} \overrightarrow{AC}.$

   Therefore, $MN \parallel AC$.

2. Ratio of lengths:

   • $AC = 9b$,
   • $MN = \frac{9b}{2}$.

   The ratio is: $\frac{AC}{MN} = \frac{9b}{\frac{9b}{2}} = 2.$

Final Answer: $MN \parallel AC \quad \text{and} \quad AC = 2 \cdot MN.$

Would you like further clarification or details?
Here are 5 related questions to consider:

1. What are the properties of midpoints in geometry?
2. How do we prove two lines are parallel using vectors?
3. What is the significance of scalar multiples in vector geometry?
4. How can we use midpoints to divide triangles into smaller similar triangles?
5. How would the solution change if $M$ and $N$ were not midpoints?

Tip: In geometry problems, always draw and label your vectors clearly to avoid confusion.