Surface area in context (SAT). Game on.

Use $SA = 2lw + 2lh + 2wh$ with $l = 6$, $w = 4$, and $h = 5$. Then $SA = 2(6)(4) + 2(6)(5) + 2(4)(5) = 48 + 60 + 40 = 148$ square centimeters.

Use $SA = 2lw + 2lh + 2wh$ with $l = 8$, $w = 3$, and $h = 4$. Then $SA = 2(8)(3) + 2(8)(4) + 2(3)(4) = 48 + 64 + 24 = 136$ for a closed box. The open top removes one $8 \times 3$ face, so subtract $24$ to get $136 - 24 = 112$ square inches.

A closed cylinder has two circular bases and a lateral surface: $SA = 2\pi r^2 + 2\pi rh$. With $r = 3$ and $h = 5$, $SA = 2\pi(9) + 2\pi(3)(5) = 18\pi + 30\pi = 48\pi$ square centimeters.

An open can has one circular base and a lateral surface: $SA = \pi r^2 + 2\pi rh$. With $r = 2$ and $h = 6$, $SA = \pi(4) + 2\pi(2)(6) = 4\pi + 24\pi = 28\pi$ square inches.

The base area is $6^2 = 36$. The four congruent triangular faces each have area $\dfrac{1}{2}(6)(5) = 15$, for a total of $4 \times 15 = 60$. The surface area is $36 + 60 = 96$ square centimeters.

With both ends open, only the lateral surface is exterior: $SA = 2\pi rh$. With $r = 2$ and $h = 10$, $SA = 2\pi(2)(10) = 40\pi$ square centimeters.

Without a floor, only the four triangular faces are counted. Along the $10$-foot sides: $2 \times \dfrac{1}{2}(10)(6) = 60$. Along the $4$-foot sides: $2 \times \dfrac{1}{2}(4)(5) = 20$. The total is $60 + 20 = 80$ square feet.

The two triangular faces have total area $2 \times \dfrac{1}{2}(3)(4) = 12$. The three rectangular faces have areas $3(10) = 30$, $4(10) = 40$, and $5(10) = 50$, where $5$ is the hypotenuse. The total is $12 + 30 + 40 + 50 = 132$ square inches.