A cuboid carton is open at the top with length l = 50 cm, width w = 30 cm, and height h = 25 cm. Calculate the outer surface area of the carton.

The total surface area of a closed cuboid is $2(lw + lh + wh)$. Since the top is open, we subtract the area of the top face ($lw$).
The formula for the outer surface area (OSA) is:

$$OSA = lw + 2(lh + wh)$$

$$OSA = 1500 \text{ cm}^2 + 2(1250 \text{ cm}^2 + 750 \text{ cm}^2)$$ $$OSA = 1500 \text{ cm}^2 + 2(2000 \text{ cm}^2)$$ $$OSA = 1500 \text{ cm}^2 + 4000 \text{ cm}^2$$ $$OSA = 5500 \text{ cm}^2$$

The outer surface area of the carton is $\boxed{\text{5500 cm}^2}$.

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Given length $L_b = 60$ cm, width $W_b = 40$ cm, and height $H_b = 30$ cm.
The formula for the total surface area (TSA) is:

$$TSA = 2(L_b W_b + L_b H_b + W_b H_b)$$

$$TSA = 2(2400 \text{ cm}^2 + 1800 \text{ cm}^2 + 1200 \text{ cm}^2)$$ $$TSA = 2(5400 \text{ cm}^2)$$ $$TSA = 10800 \text{ cm}^2$$

The surface area of each board (for one box) is $\boxed{\text{10800 cm}^2}$.

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When the tank is full, the water touches the base and the four side walls. It does not touch the top.
The formula for the area in contact with water is:

$$Area = l_t w_t + 2(l_t h_t + w_t h_t)$$

$$Area = 11.25 \text{ m}^2 + 2(4.5 \text{ m}^2 + 2.5 \text{ m}^2)$$ $$Area = 11.25 \text{ m}^2 + 2(7.0 \text{ m}^2)$$ $$Area = 11.25 \text{ m}^2 + 14.0 \text{ m}^2$$ $$Area = 25.25 \text{ m}^2$$

The surface area of the tank in contact with water is $\boxed{\text{25.25 m}^2}$.

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Given total surface area (TSA) = 126 cm$^2$, length $l_c = 6$ cm, and height $h_c = 3$ cm. Let the width be $w_c$.
The formula for the total surface area (TSA) is:

$$TSA = 2(l_c w_c + l_c h_c + w_c h_c)$$

$$126 = 2(6w_c + 18 + 3w_c)$$ $$126 = 2(9w_c + 18)$$

Divide both sides by 2:

$$\frac{126}{2} = 9w_c + 18$$ $$63 = 9w_c + 18$$

Subtract 18 from both sides:

$$63 - 18 = 9w_c$$ $$45 = 9w_c$$

Divide by 9:

$$w_c = \frac{45}{9}$$ $$w_c = 5 \text{ cm}$$

The width of the cuboid is $\boxed{\text{5 cm}}$.