Calculate the cross-sectional areas and velocities in both pipe sections.

The volumetric flow rate $Q$ is given as $3.2 \text{ m}^3/\text{min}$.

$$Q = 3.2 \frac{\text{m}^3}{\text{min}} \times \frac{1 \text{ min}}{60 \text{ s}} = 0.05333 \frac{\text{m}^3}{\text{s}}$$

The diameters are $D_1 = 120 \text{ mm} = 0.12 \text{ m}$ and $D_2 = 180 \text{ mm} = 0.18 \text{ m}$.
The cross-sectional areas are:

$$A_1 = \frac{\pi}{4} D_1^2 = \frac{\pi}{4} (0.12 \text{ m})^2 = 0.01131 \text{ m}^2$$ $$A_2 = \frac{\pi}{4} D_2^2 = \frac{\pi}{4} (0.18 \text{ m})^2 = 0.02545 \text{ m}^2$$

The velocities are:

$$V_1 = \frac{Q}{A_1} = \frac{0.05333 \text{ m}^3/\text{s}}{0.01131 \text{ m}^2} = 4.716 \text{ m/s}$$ $$V_2 = \frac{Q}{A_2} = \frac{0.05333 \text{ m}^3/\text{s}}{0.02545 \text{ m}^2} = 2.096 \text{ m/s}$$

The formula for head loss due to sudden enlargement is:

$$h_L = \frac{(V_1 - V_2)^2}{2g}$$

Using $g = 9.81 \text{ m/s}^2$:

$$h_L = \frac{(4.716 \text{ m/s} - 2.096 \text{ m/s})^2}{2 \times 9.81 \text{ m/s}^2} = \frac{(2.620 \text{ m/s})^2}{19.62 \text{ m/s}^2} = \frac{6.864 \text{ m}^2/\text{s}^2}{19.62 \text{ m/s}^2} = 0.3498 \text{ m}$$

The head loss due to the abrupt change in pipe diameter is $\boxed{\text{0.350 m}}$.

For a horizontal pipe ($z_1 = z_2$), Bernoulli's equation with head loss is:

$$\frac{P_1}{\rho g} + \frac{V_1^2}{2g} = \frac{P_2}{\rho g} + \frac{V_2^2}{2g} + h_L$$

Rearranging to find the pressure difference $P_1 - P_2$:

$$P_1 - P_2 = \rho g \left( \frac{V_2^2 - V_1^2}{2g} + h_L \right)$$

Using $\rho = 1000 \text{ kg/m}^3$:

$$P_1 - P_2 = 1000 \text{ kg/m}^3 \left( \frac{(2.096 \text{ m/s})^2 - (4.716 \text{ m/s})^2}{2 \times 9.81 \text{ m/s}^2} + 0.3498 \text{ m} \right)$$ $$P_1 - P_2 = 1000 \text{ kg/m}^3 \left( \frac{4.393 \text{ m}^2/\text{s}^2 - 22.24 \text{ m}^2/\text{s}^2}{19.62 \text{ m/s}^2} + 0.3498 \text{ m} \right)$$ $$P_1 - P_2 = 1000 \text{ kg/m}^3 \left( \frac{-17.847 \text{ m}^2/\text{s}^2}{19.62 \text{ m/s}^2} + 0.3498 \text{ m} \right)$$ $$P_1 - P_2 = 1000 \text{ kg/m}^3 (-0.9096 \text{ m} + 0.3498 \text{ m})$$ $$P_1 - P_2 = 1000 \text{ kg/m}^3 (-0.5598 \text{ m}) = -559.8 \text{ kg/(m}\cdot\text{s}^2) = -559.8 \text{ Pa}$$

The pressure difference between the two sections is $\boxed{\text{-560 Pa}}$. (A negative value indicates $P_2 > P_1$).

If the transition is frictionless, $h_L = 0$.

$$\frac{P_1}{\rho g} + \frac{V_1^2}{2g} = \frac{P_2}{\rho g} + \frac{V_2^2}{2g}$$

Rearranging for $P_1 - P_2$:

$$P_1 - P_2 = \rho g \left( \frac{V_2^2 - V_1^2}{2g} \right) = \frac{\rho}{2} (V_2^2 - V_1^2)$$ $$P_1 - P_2 = \frac{1000 \text{ kg/m}^3}{2} ((2.096 \text{ m/s})^2 - (4.716 \text{ m/s})^2)$$ $$P_1 - P_2 = 500 \text{ kg/m}^3 (4.393 \text{ m}^2/\text{s}^2 - 22.24 \text{ m}^2/\text{s}^2)$$ $$P_1 - P_2 = 500 \text{ kg/m}^3 (-17.847 \text{ m}^2/\text{s}^2) = -8923.5 \text{ Pa}$$

The change in pressure for a frictionless transition is $\boxed{\text{-8920 Pa}}$.

The Darcy-Weisbach formula is:

$$h_f = \frac{f L V^2}{2 g D}$$

Using $g = 9.81 \text{ m/s}^2$:

$$h_f = \frac{0.006 \times 80 \text{ m} \times (3.0 \text{ m/s})^2}{2 \times 9.81 \text{ m/s}^2 \times 0.25 \text{ m}}$$ $$h_f = \frac{0.006 \times 80 \times 9}{4.905} = \frac{4.32}{4.905} = 0.8807 \text{ m}$$

The head loss due to friction using Darcy-Weisbach is $\boxed{\text{0.881 m}}$.

First, calculate the hydraulic radius $R$ for a circular pipe:

$$R = \frac{D}{4} = \frac{0.25 \text{ m}}{4} = 0.0625 \text{ m}$$

Chezy's formula is $V = C \sqrt{R S}$, where $S = \frac{h_f}{L}$.
Rearranging to solve for $h_f$:

$$V^2 = C^2 R \frac{h_f}{L}$$ $$h_f = \frac{V^2 L}{C^2 R}$$ $$h_f = \frac{(3.0 \text{ m/s})^2 \times 80 \text{ m}}{(50)^2 \times 0.0625 \text{ m}}$$ $$h_f = \frac{9 \times 80}{2500 \times 0.0625} = \frac{720}{156.25} = 4.608 \text{ m}$$

The head loss due to friction using Chezy's formula is $\boxed{\text{4.61 m}}$.