Question 1170318
Let $h$ be the altitude of the right circular cylinder and $r$ be the radius of its base. We are given $h = 12$ inches and $r = 12$ inches.

**Inscribed Triangular Prism**

For the inscribed triangular prism, the base is an equilateral triangle inscribed in a circle of radius $r$. Let $s$ be the side length of the equilateral triangle.
The relationship between $s$ and $r$ is $r = \frac{s}{\sqrt{3}}$, so $s = r\sqrt{3}$.
Since $r = 12$, $s = 12\sqrt{3}$.

The area of the equilateral triangle is $A_{inscribed} = \frac{\sqrt{3}}{4} s^2 = \frac{\sqrt{3}}{4} (12\sqrt{3})^2 = \frac{\sqrt{3}}{4} (144 \cdot 3) = 108\sqrt{3}$.

The lateral surface area of the inscribed prism is $3 \cdot s \cdot h = 3 \cdot 12\sqrt{3} \cdot 12 = 432\sqrt{3}$.

The total surface area of the inscribed prism is $2 A_{inscribed} + 3sh = 2(108\sqrt{3}) + 432\sqrt{3} = 216\sqrt{3} + 432\sqrt{3} = 648\sqrt{3}$.

**Circumscribed Triangular Prism**

For the circumscribed triangular prism, the base is an equilateral triangle circumscribed about a circle of radius $r$.
The relationship between the side length $S$ of the equilateral triangle and the radius $r$ is $r = \frac{S}{2\sqrt{3}}$, so $S = 2\sqrt{3} r$.
Since $r = 12$, $S = 2\sqrt{3} \cdot 12 = 24\sqrt{3}$.

The area of the equilateral triangle is $A_{circumscribed} = \frac{\sqrt{3}}{4} S^2 = \frac{\sqrt{3}}{4} (24\sqrt{3})^2 = \frac{\sqrt{3}}{4} (576 \cdot 3) = 432\sqrt{3}$.

The lateral surface area of the circumscribed prism is $3 \cdot S \cdot h = 3 \cdot 24\sqrt{3} \cdot 12 = 864\sqrt{3}$.

The total surface area of the circumscribed prism is $2 A_{circumscribed} + 3Sh = 2(432\sqrt{3}) + 864\sqrt{3} = 864\sqrt{3} + 864\sqrt{3} = 1728\sqrt{3}$.

**Total Surface Areas**

The total surface area of the inscribed prism is $648\sqrt{3}$ square inches.
The total surface area of the circumscribed prism is $1728\sqrt{3}$ square inches.

Final Answer: The final answer is $\boxed{648\sqrt{3}, 1728\sqrt{3}}$