Calculate which container has the largest volume. Give the volume to the nearest litre.

For the rectangular container:
Length ($l$) = $80 \text{ cm} = 0.8 \text{ m}$
Width ($w$) = $80 \text{ cm} = 0.8 \text{ m}$
Height ($h$) = $2 \text{ m}$

For the cylindrical container:
Diameter ($D$) = $1 \text{ m}$
Radius ($r$) = $\frac{D}{2} = \frac{1 \text{ m}}{2} = 0.5 \text{ m}$
Height ($h$) = $2 \text{ m}$

$$V_{\text{rectangular}} = l \times w \times h$$ $$V_{\text{rectangular}} = 0.8 \text{ m} \times 0.8 \text{ m} \times 2 \text{ m}$$ $$V_{\text{rectangular}} = 1.28 \text{ m}^3$$

Given $1000 \text{ cm}^3 = 1 \text{ l}$, and $1 \text{ m}^3 = (100 \text{ cm})^3 = 1,000,000 \text{ cm}^3$.
Therefore, $1 \text{ m}^3 = \frac{1,000,000}{1000} \text{ l} = 1000 \text{ l}$.

$$V_{\text{rectangular}} = 1.28 \text{ m}^3 \times 1000 \frac{\text{l}}{\text{m}^3}$$ $$V_{\text{rectangular}} = 1280 \text{ l}$$

$$V_{\text{cylindrical}} = \pi r^2 h$$ $$V_{\text{cylindrical}} = \pi \times (0.5 \text{ m})^2 \times 2 \text{ m}$$ $$V_{\text{cylindrical}} = \pi \times 0.25 \text{ m}^2 \times 2 \text{ m}$$ $$V_{\text{cylindrical}} = 0.5\pi \text{ m}^3$$

Using $\pi \approx 3.14159$:

$$V_{\text{cylindrical}} \approx 0.5 \times 3.14159 \text{ m}^3$$ $$V_{\text{cylindrical}} \approx 1.570795 \text{ m}^3$$

$$V_{\text{cylindrical}} \approx 1.570795 \text{ m}^3 \times 1000 \frac{\text{l}}{\text{m}^3}$$ $$V_{\text{cylindrical}} \approx 1570.795 \text{ l}$$ $$V_{\text{cylindrical}} \approx 1571 \text{ l}$$

$1571 \text{ l} > 1280 \text{ l}$.
The cylindrical container has the largest volume.

The container with the largest volume is the cylindrical container, with a volume of $\boxed{\text{1571 l}}$.

$$SA_{\text{rectangular}} = 2(lw + lh + wh)$$ $$SA_{\text{rectangular}} = 2((0.8 \text{ m} \times 0.8 \text{ m}) + (0.8 \text{ m} \times 2 \text{ m}) + (0.8 \text{ m} \times 2 \text{ m}))$$ $$SA_{\text{rectangular}} = 2(0.64 \text{ m}^2 + 1.6 \text{ m}^2 + 1.6 \text{ m}^2)$$ $$SA_{\text{rectangular}} = 2(3.84 \text{ m}^2)$$ $$SA_{\text{rectangular}} = 7.68 \text{ m}^2$$

$$SA_{\text{cylindrical}} = 2\pi r^2 + 2\pi rh$$ $$SA_{\text{cylindrical}} = 2\pi (0.5 \text{ m})^2 + 2\pi (0.5 \text{ m})(2 \text{ m})$$ $$SA_{\text{cylindrical}} = 2\pi (0.25 \text{ m}^2) + 2\pi (1 \text{ m}^2)$$ $$SA_{\text{cylindrical}} = 0.5\pi \text{ m}^2 + 2\pi \text{ m}^2$$ $$SA_{\text{cylindrical}} = 2.5\pi \text{ m}^2$$

Using $\pi \approx 3.14159$:

$$SA_{\text{cylindrical}} \approx 2.5 \times 3.14159 \text{ m}^2$$ $$SA_{\text{cylindrical}} \approx 7.853975 \text{ m}^2$$

The surface area of the rectangular container is $\boxed{\text{7.68 m}^2}$.
The surface area of the cylindrical container is $\boxed{\text{7.85 m}^2}$ (rounded to two decimal places).

Length of sheeting = Total Surface Area / Width of sheet

$$L_{\text{rectangular}} = \frac{SA_{\text{rectangular}}}{\text{width of sheet}}$$ $$L_{\text{rectangular}} = \frac{7.68 \text{ m}^2}{2 \text{ m}}$$ $$L_{\text{rectangular}} = 3.84 \text{ m}$$

$$L_{\text{cylindrical}} = \frac{SA_{\text{cylindrical}}}{\text{width of sheet}}$$ $$L_{\text{cylindrical}} = \frac{7.853975 \text{ m}^2}{2 \text{ m}}$$ $$L_{\text{cylindrical}} \approx 3.9269875 \text{ m}$$ $$L_{\text{cylindrical}} \approx 3.93 \text{ m}$$

The length of sheeting required for the rectangular container is $\boxed{\text{3.84 m}}$.
The length of sheeting required for the cylindrical container is $\boxed{\text{3.93 m}}$ (rounded to two decimal places).