Given sigma_u = 480 MPa and FOS = 4:

The allowable working stress ($\sigma_w$) is determined by dividing the ultimate stress ($\sigma_u$) by the factor of safety (FOS).

$$\sigma_w = \frac{\sigma_u}{\text{FOS}}$$

Given $\sigma_u = 480 \, \text{MPa}$ and $\text{FOS} = 4$:

$$\sigma_w = \frac{480 \, \text{MPa}}{4}$$ $$\sigma_w = 120 \, \text{MPa}$$

The allowable working stress is $\boxed{\text{120 MPa}}$.

The tensile load ($P$) is $400 \, \text{kN}$, which is $400 \times 10^3 \, \text{N}$. The allowable working stress ($\sigma_w$) is $120 \, \text{MPa}$, which is $120 \times 10^6 \, \text{Pa}$ (N/m$^2$).
The area ($A$) is calculated using the formula $\sigma_w = \frac{P}{A}$, so $A = \frac{P}{\sigma_w}$.

$$A = \frac{400 \times 10^3 \, \text{N}}{120 \times 10^6 \, \text{Pa}}$$ $$A = \frac{400}{120000} \, \text{m}^2$$ $$A = \frac{1}{300} \, \text{m}^2 \approx 0.003333 \, \text{m}^2$$

The tube is hollow with an inside diameter ($d_i$) of $100 \, \text{mm} = 0.1 \, \text{m}$. The area of a hollow tube is given by:

$$A = \frac{\pi}{4} (d_o^2 - d_i^2)$$

We need to solve for the outside diameter ($d_o$). Rearrange the formula:

$$\frac{4A}{\pi} = d_o^2 - d_i^2$$ $$d_o^2 = \frac{4A}{\pi} + d_i^2$$ $$d_o = \sqrt{\frac{4A}{\pi} + d_i^2}$$

Substitute the values for $A$ and $d_i$:

$$d_o = \sqrt{\frac{4 \times (1/300) \, \text{m}^2}{\pi} + (0.1 \, \text{m})^2}$$ $$d_o = \sqrt{\frac{1}{75\pi} + 0.01} \, \text{m}$$ $$d_o = \sqrt{0.00424413 + 0.01} \, \text{m}$$ $$d_o = \sqrt{0.01424413} \, \text{m}$$ $$d_o \approx 0.119348 \, \text{m}$$

Convert the outside diameter to millimeters:

$$d_o \approx 0.119348 \times 1000 \, \text{mm}$$ $$d_o \approx 119.35 \, \text{mm}$$

The outside diameter of the tube is $\boxed{\text{119.35 mm}}$.