Calculate the radius of the cylindrical tank.

The diameter is $5.6 \text{ m}$, so the radius $r$ is half of the diameter.

$$r = \frac{5.6 \text{ m}}{2} = 2.8 \text{ m}$$

The formula for the volume of a cylinder is $V = \pi r^2 h$.
Given $h_c = 2.5 \text{ m}$. We will use $\pi = \frac{22}{7}$.

$$V_c = \pi (2.8 \text{ m})^2 (2.5 \text{ m})$$ $$V_c = \frac{22}{7} \times (2.8 \text{ m}) \times (2.8 \text{ m}) \times (2.5 \text{ m})$$ $$V_c = 22 \times (0.4 \text{ m}) \times (2.8 \text{ m}) \times (2.5 \text{ m})$$ $$V_c = 8.8 \text{ m} \times 2.8 \text{ m} \times 2.5 \text{ m}$$ $$V_c = 24.64 \text{ m}^2 \times 2.5 \text{ m}$$ $$V_c = 61.6 \text{ m}^3$$

The volume of water transferred to the rectangular tank is the same as the volume in the cylindrical tank.
The formula for the volume of a rectangular tank is $V = L \times W \times h$.
Given $L = 5.5 \text{ m}$ and $W = 4 \text{ m}$. Let $h_r$ be the height of the water in the rectangular tank.

$$V_r = (5.5 \text{ m}) \times (4 \text{ m}) \times h_r$$ $$V_r = 22 h_r \text{ m}^2$$

Since $V_c = V_r$:

$$61.6 \text{ m}^3 = 22 h_r \text{ m}^2$$ $$h_r = \frac{61.6 \text{ m}^3}{22 \text{ m}^2}$$ $$h_r = 2.8 \text{ m}$$

The height of the water in the rectangular tank was $\boxed{2.8 \text{ m}}$.

: Half Marathon Speed

The race is $21 \text{ km}$ long.

$$1 \text{ km} = 1000 \text{ m}$$ $$D = 21 \text{ km} \times \frac{1000 \text{ m}}{1 \text{ km}} = 21000 \text{ m}$$

Peter ran the race in $1 \text{ hour } 40 \text{ minutes}$.

$$1 \text{ hour} = 60 \text{ minutes}$$

Total minutes = $60 \text{ minutes} + 40 \text{ minutes} = 100 \text{ minutes}$$

$$1 \text{ minute} = 60 \text{ seconds}$$

Total seconds = $100 \text{ minutes} \times \frac{60 \text{ seconds}}{1 \text{ minute}} = 6000 \text{ seconds}$$

The formula for speed is $S = \frac{D}{T}$.

$$S = \frac{21000 \text{ m}}{6000 \text{ s}}$$ $$S = \frac{21}{6} \text{ m/s}$$ $$S = 3.5 \text{ m/s}$$

Peter's speed was $\boxed{3.5 \text{ m/s}}$.