To solve this problem, we need to break it down step by step, using trigonometry and the properties of triangles.

Step 1: Find the third angle of the triangle.

We know the sum of the angles in a triangle is always $180^\circ$. Therefore, angle $C$ can be found as follows: $\angle C = 180^\circ - \angle A - \angle B = 180^\circ - 45^\circ - 60^\circ = 75^\circ.$

Step 2: Use the Law of Sines to find side lengths.

The Law of Sines states: $\frac{AB}{\sin C} = \frac{BC}{\sin A} = \frac{AC}{\sin B}.$ We are given that $AB = 5$, and we know the angles $A = 45^\circ$, $B = 60^\circ$, and $C = 75^\circ$.

Using the Law of Sines to find $BC$ and $AC$:

$\frac{AB}{\sin C} = \frac{BC}{\sin A} \quad \Rightarrow \quad \frac{5}{\sin 75^\circ} = \frac{BC}{\sin 45^\circ}.$

Since $\sin 45^\circ = \frac{\sqrt{2}}{2}$, this becomes: $\frac{5}{\sin 75^\circ} = \frac{BC}{\frac{\sqrt{2}}{2}} \quad \Rightarrow \quad BC = \frac{5 \cdot \sqrt{2}}{2 \cdot \sin 75^\circ}.$

Using a calculator or known values, $\sin 75^\circ \approx 0.9659$, so: $BC \approx \frac{5 \cdot \sqrt{2}}{2 \cdot 0.9659} \approx \frac{5 \cdot 1.414}{1.9318} \approx \frac{7.07}{1.9318} \approx 3.66 \, \text{units}.$

Now, use the Law of Sines to find $AC$:

$\frac{AB}{\sin C} = \frac{AC}{\sin B} \quad \Rightarrow \quad \frac{5}{\sin 75^\circ} = \frac{AC}{\sin 60^\circ}.$

Since $\sin 60^\circ = \frac{\sqrt{3}}{2}$, this becomes: $\frac{5}{\sin 75^\circ} = \frac{AC}{\frac{\sqrt{3}}{2}} \quad \Rightarrow \quad AC = \frac{5 \cdot \sqrt{3}}{2 \cdot \sin 75^\circ}.$

Substituting the value for $\sin 75^\circ$, we get: $AC \approx \frac{5 \cdot 1.732}{2 \cdot 0.9659} \approx \frac{8.66}{1.9318} \approx 4.48 \, \text{units}.$

Step 3: Apply the section formula for $AP$.

We are told that point $P$ divides side $BC$ such that $\frac{BP}{PC} = \frac{1}{2}$. This means $P$ divides $BC$ in a 1:2 ratio, so:

$\frac{BP}{PC} = \frac{1}{2} \quad \Rightarrow \quad BP = \frac{BC}{3}, \quad PC = \frac{2BC}{3}.$

Now, using the section formula for the length $AP$, we can express the length of $AP$ as:

$AP^2 = \frac{AB^2 \cdot PC + AC^2 \cdot BP}{BC}.$

Substituting the values we have:

$AP^2 = \frac{5^2 \cdot \frac{2BC}{3} + 4.48^2 \cdot \frac{BC}{3}}{BC}.$

Simplifying:

$AP^2 = \frac{25 \cdot \frac{2}{3} \cdot BC + 20.06 \cdot \frac{1}{3} \cdot BC}{BC}.$

Factoring out $BC$ and simplifying:

$AP^2 = \frac{BC}{3} \left( 50 + 20.06 \right) = \frac{BC}{3} \cdot 70.06.$

Since $BC \approx 3.66$, we get:

$AP^2 \approx \frac{3.66}{3} \cdot 70.06 \approx 1.22 \cdot 70.06 \approx 85.47.$

Therefore, $AP \approx \sqrt{85.47} \approx 9.25$.

Step 4: Find $AP^6$.

Finally, we need to find $AP^6$:

$AP^6 = (9.25)^6 \approx 7375.46.$

Thus, the value of $AP^6$ is approximately 7375.46.

Would you like any more details or clarifications on any of the steps? Here are some related questions to explore further:

1. How does the Law of Sines work in non-right triangles?
2. What is the section formula and how can it be applied in different situations?
3. How can trigonometric identities help simplify expressions in triangle problems?
4. What are other methods to solve for distances in triangles, such as using the Law of Cosines?
5. How do fractional exponents relate to rational expressions in geometry?

Tip: When solving triangle problems, always try to visualize the situation and break it down into smaller steps, especially using laws like the Law of Sines and Law of Cosines!