A hollow circular shaft has inside and outside diameters of 30 mm and 40 mm, respectively. The material has a shear modulus of 60 kN/mm². Determine the maximum stress in the shaft when it is subjected to a twist of 5° in a length of 2.5 m. What is the maximum torque the shaft can carry?

The power ($P$) transmitted by a shaft is given by $P = \frac{2 \pi N T}{60}$, where $N$ is the rotational speed in rpm.
Given $P = 2.2 \text{ MW} = 2.2 \times 10^6 \text{ W}$ and $N = 60 \text{ rpm}$.

$$T = \frac{P \times 60}{2 \pi N}$$ $$T = \frac{2.2 \times 10^6 \text{ W} \times 60}{2 \pi \times 60 \text{ rpm}}$$ $$T = \frac{2.2 \times 10^6}{2 \pi} \text{ N}\cdot\text{m}$$ $$T \approx 350140.85 \text{ N}\cdot\text{m}$$

For a solid shaft, the maximum shear stress ($\tau_{max}$) is given by $\tau_{max} = \frac{16 T}{\pi D_s^3}$.
Given $\tau_{max} = 85 \text{ MPa} = 85 \times 10^6 \text{ N/m}^2$.

$$D_s^3 = \frac{16 T}{\pi \tau_{max}}$$ $$D_s^3 = \frac{16 \times 350140.85 \text{ N}\cdot\text{m}}{\pi \times 85 \times 10^6 \text{ N/m}^2}$$ $$D_s^3 = \frac{5602253.6}{\pi \times 85 \times 10^6} \text{ m}^3$$ $$D_s^3 \approx 0.02098 \text{ m}^3$$ $$D_s = (0.02098)^{1/3} \text{ m}$$ $$D_s \approx 0.2758 \text{ m} = 275.8 \text{ mm}$$

The required diameter of the solid shaft is $\boxed{\text{275.8 mm}}$.

For a hollow shaft, the maximum shear stress ($\tau_{max}$) is given by $\tau_{max} = \frac{16 T}{\pi D_e^3 (1 - k^4)}$, where $k = \frac{D_i}{D_e}$ is the ratio of internal to external diameter.
Given $3D_e = 4D_i$, which means $k = \frac{D_i}{D_e} = \frac{3}{4} = 0.75$.
The hollow shaft transmits the same torque $T \approx 350140.85 \text{ N}\cdot\text{m}$ and has the same allowable shear stress $\tau_{max} = 85 \times 10^6 \text{ N/m}^2$.

$$D_e^3 = \frac{16 T}{\pi \tau_{max} (1 - k^4)}$$

First, calculate $1 - k^4$:

$$1 - k^4 = 1 - (0.75)^4 = 1 - 0.31640625 = 0.68359375$$

Now, substitute the values:

$$D_e^3 = \frac{16 \times 350140.85 \text{ N}\cdot\text{m}}{\pi \times 85 \times 10^6 \text{ N/m}^2 \times 0.68359375}$$ $$D_e^3 = \frac{5602253.6}{\pi \times 85 \times 10^6 \times 0.68359375} \text{ m}^3$$ $$D_e^3 \approx 0.03068 \text{ m}^3$$ $$D_e = (0.03068)^{1/3} \text{ m}$$ $$D_e \approx 0.3130 \text{ m}$$

$$D_i = k \times D_e = 0.75 \times 0.3130 \text{ m}$$ $$D_i \approx 0.23475 \text{ m}$$

The weight per meter length is proportional to the cross-sectional area for the same material.
Area of solid shaft ($A_s$):

$$A_s = \frac{\pi}{4} D_s^2 = \frac{\pi}{4} (0.2758 \text{ m})^2$$ $$A_s \approx 0.05973 \text{ m}^2$$

Area of hollow shaft ($A_h$):

$$A_h = \frac{\pi}{4} (D_e^2 - D_i^2) = \frac{\pi}{4} ((0.3130 \text{ m})^2 - (0.23475 \text{ m})^2)$$ $$A_h = \frac{\pi}{4} (0.097969 - 0.0551075625) \text{ m}^2$$ $$A_h = \frac{\pi}{4} (0.0428614375) \text{ m}^2$$ $$A_h \approx 0.03367 \text{ m}^2$$

$$\text{Percentage Saving} = \frac{A_s - A_h}{A_s} \times 100\%$$ $$\text{Percentage Saving} = \frac{0.05973 \text{ m}^2 - 0.03367 \text{ m}^2}{0.05973 \text{ m}^2} \times 100\%$$ $$\text{Percentage Saving} = \frac{0.02606}{0.05973} \times 100\%$$ $$\text{Percentage Saving} \approx 43.63\%$$

The percentage saving in weight per meter length is $\boxed{\text{43.63\%}}$.