Calculate the semi-perimeter of triangle PQR.

The side lengths are $PQ = c = 7.8$ cm, $PR = b = 6.6$ cm, and $QR = a = 5.9$ cm.
The semi-perimeter $s$ is given by:

$$s = \frac{a+b+c}{2}$$ $$s = \frac{5.9 + 6.6 + 7.8}{2}$$ $$s = \frac{20.3}{2}$$ $$s = 10.15 \text{ cm}$$

The area $K$ is given by:

$$K = \sqrt{s(s-a)(s-b)(s-c)}$$ $$K = \sqrt{10.15(10.15-5.9)(10.15-6.6)(10.15-7.8)}$$ $$K = \sqrt{10.15(4.25)(3.55)(2.35)}$$ $$K = \sqrt{359.0068125}$$ $$K \approx 18.94748 \text{ cm}^2$$

a) Step 3: Calculate the radius of the circumcircle.
The radius $R$ of the circumcircle is given by the formula:

$$R = \frac{abc}{4K}$$ $$R = \frac{5.9 \times 6.6 \times 7.8}{4 \times 18.94748}$$ $$R = \frac{304.788}{75.78992}$$ $$R \approx 4.0215 \text{ cm}$$

Correct to one decimal place:

$$R \approx 4.0 \text{ cm}$$

The radius of the circle is $\boxed{4.0 \text{ cm}}$.

b) Step 4: Calculate the angles of the triangle using the Cosine Rule.
For angle P (opposite side $a=5.9$ cm):

$$\cos P = \frac{b^2 + c^2 - a^2}{2bc}$$ $$\cos P = \frac{6.6^2 + 7.8^2 - 5.9^2}{2 \times 6.6 \times 7.8}$$ $$\cos P = \frac{43.56 + 60.84 - 34.81}{102.96}$$ $$\cos P = \frac{69.59}{102.96} \approx 0.67599$$ $$P = \arccos(0.67599) \approx 47.47^\circ$$

For angle Q (opposite side $b=6.6$ cm):

$$\cos Q = \frac{a^2 + c^2 - b^2}{2ac}$$ $$\cos Q = \frac{5.9^2 + 7.8^2 - 6.6^2}{2 \times 5.9 \times 7.8}$$ $$\cos Q = \frac{34.81 + 60.84 - 43.56}{92.04}$$ $$\cos Q = \frac{52.09}{92.04} \approx 0.56595$$ $$Q = \arccos(0.56595) \approx 55.53^\circ$$

For angle R (opposite side $c=7.8$ cm):

$$\cos R = \frac{a^2 + b^2 - c^2}{2ab}$$ $$\cos R = \frac{5.9^2 + 6.6^2 - 7.8^2}{2 \times 5.9 \times 6.6}$$ $$\cos R = \frac{34.81 + 43.56 - 60.84}{77.88}$$ $$\cos R = \frac{17.53}{77.88} \approx 0.22509$$ $$R = \arccos(0.22509) \approx 77.00^\circ$$

The angles of the triangle are $\boxed{P \approx 47.5^\circ, Q \approx 55.5^\circ, R \approx 77.0^\circ}$.

c) Step 5: Calculate the area of the shaded region.
The area of the circle is $A_{circle} = \pi R^2$. Using the more precise value of $R \approx 4.0215$ cm:

$$A_{circle} = \pi (4.0215)^2$$ $$A_{circle} = \pi (16.17246225)$$ $$A_{circle} \approx 50.806 \text{ cm}^2$$

The area of