The tetrahedron has four faces, each of which is an equilateral triangle with sidelength $\displaystyle s$. Since the total surface area is 444, each triangle has area one fourth of this, or 111. To find $\displaystyle s$, set $\displaystyle A = 111$ in the formula for the area of an equilateral triangle:

$\displaystyle \frac{s^{2} \sqrt{3} }{4} =A$

$\displaystyle \frac{s^{2} \sqrt{3} }{4} =111$

$\displaystyle s^{2} \sqrt{3} =444$

$\displaystyle s^{2}=\frac{444}{ \sqrt{3} } \approx \frac{444}{ 1.732 } \approx 256.34$

$\displaystyle s\approx \sqrt{256.34} \approx 16.0$

We are looking for the height of the pyramid.

The base is an equilateral triangle with sidelength 4, so its area can be calculated as follows:

$\displaystyle B = \frac{s^{2} \sqrt{3} }{4}$

$\displaystyle B = \frac{4^{2} \sqrt{3} }{4}$

$\displaystyle B = \frac{16 \sqrt{3} }{4}$

$\displaystyle B = 4\sqrt{3}$

The height $\displaystyle h$ of a pyramid can be calculated using the fomula

$\displaystyle V = \frac{1}{3}Bh$

We set $\displaystyle V = 25$ and $\displaystyle B = 4\sqrt{3}$ and solve for $\displaystyle h$:

$\displaystyle \frac{1}{3} \cdot 4\sqrt{3} \cdot h = 25$

$\displaystyle h = \frac{25 \cdot 3}{4 \sqrt{3}} = \frac{75}{4 \sqrt{3}} \approx \frac{75}{4 \cdot 1.7321} \approx 10.8$

The height of the pyramid is $\displaystyle h = 8$. The base is an equilateral triangle with sidelength 4, so its area can be calculated as follows:

$\displaystyle B = \frac{s^{2} \sqrt{3} }{4}$

$\displaystyle B = \frac{4^{2} \sqrt{3} }{4}$

$\displaystyle B = \frac{16 \sqrt{3} }{4}$

$\displaystyle B = 4\sqrt{3}$

The volume of a pyramid can be calculated using the fomula

$\displaystyle V = \frac{1}{3}Bh$

$\displaystyle V = \frac{1}{3} \cdot 4\sqrt{3} \cdot 8 = \frac{32\sqrt{3} }{3} \approx \frac{32 \cdot 1.7321}{3} \approx 18.5$

The volume of a tetrahedron is found with the equation $\displaystyle V=\frac{a^3}{6\sqrt{2}}$, where $\displaystyle a$ represents the length of an edge of the tetrahedron.

Plug in 4 for the edge length and reduce as much as possible to find the answer:

$\displaystyle V=\frac{4^3}{6\sqrt{2}}$

$\displaystyle V=\frac{64}{6\sqrt{2}}$

$\displaystyle V=\frac{32}{3\sqrt{2}}$

The volume of the tetrahedron is $\displaystyle \frac{32}{3\sqrt{2}}$.

The area of an equilateral triangle is given by the formula

$\displaystyle A = \frac{s^{2}\sqrt{3}}{4}$

Since there are four equilateral triangles that comprise the surface of the tetrahedron, the total surface area is 

$\displaystyle SA = 4 A = 4 \cdot \frac{s^{2}\sqrt{3}}{4} = s^{2}\sqrt{3}$

Substitute $\displaystyle s = 6$:

$\displaystyle SA = s^{2}\sqrt{3} = 6^{2}\sqrt{3} = 36\sqrt{3}$ square inches.

The tetrahedron has four faces, each of which is an equilateral triangle with sidelength 7. Each face has area

$\displaystyle A = \frac{s^{2} \sqrt{3} }{4} = \frac{7^{2} \sqrt{3} }{4} = \frac{49 \sqrt{3} }{4} \approx \frac{49 \cdot 1.7321}{4} \approx 21.22$

The total surface area is four times this, or about $\displaystyle 4 \cdot 21.22 = 84.88$.

Rounded, this is 84.9.