Given: altitude (h) = 6 cm, base edge (s) = 4 cm.

Given: altitude ($h$) = 6 cm, base edge ($s$) = 4 cm.
The volume of a pyramid is $V = \frac{1}{3} \times \text{Base Area} \times h$.

Base Area ($A_b$) = $s^2 = (4 \text{ cm})^2 = 16 \text{ cm}^2$.

The volume of the pyramid is $\boxed{\text{32 cm}^3}$.

2. A circular cone has an altitude 12 cm and a base radius 10 cm. What is its volume?

Given: altitude ($h$) = 12 cm, base radius ($r$) = 10 cm.
The volume of a cone is $V = \frac{1}{3}\pi r^2 h$.

Using $\pi \approx 3.14159$:

The volume of the cone is $\boxed{\text{400\pi cm}^3 \text{ or approximately 1256.64 cm}^3}$.

3. A right circular cone has height 10 cm and circumference of the base is $12\pi$ cm. Find its volume.

Given: height ($h$) = 10 cm, circumference ($C$) = $12\pi$ cm.
The formula for circumference is $C = 2\pi r$.

Divide both sides by $2\pi$:

The volume of a cone is $V = \frac{1}{3}\pi r^2 h$.

Using $\pi \approx 3.14159$:

The volume of the cone is $\boxed{\text{120\pi cm}^3 \text{ or approximately 376.99 cm}^3}$.

4. The lateral edge of a regular tetrahedron is 6 cm. Find its total surface area and its volume.

Given: lateral edge ($a$) = 6 cm.
A regular tetrahedron has 4 equilateral triangular faces.
Area of one equilateral triangle with side $a$: $A_{face} = \frac{\sqrt{3}}{4}a^2$.
Total Surface Area ($TSA$) = $4 \times A_{face} = \sqrt{3}a^2$.
Volume ($V$) of a regular tetrahedron with edge $a$: $V = \frac{\sqrt{2}}{12}a^3$.

Using $\sqrt{3} \approx 1.732$:

Using $\sqrt{2} \approx 1.414$:

The total surface area is \boxed{\text{36\sqrt{3} cm}^2 \text{ or approximately 62.35 cm}^2} and the volume is \boxed{\text{18\sqrt{2} cm}^3 \text{ or approximately 25.45 cm}^3}.

5. A right circular conical vessel of altitude 20 cm and base radius 10 cm is kept with its vertex downwards. If one liter of water is poured into it, how high above the vertex will the level of the water be? Use $\pi = 3.14$.

Vessel altitude ($H$) = 20 cm, vessel base radius ($R$) = 10 cm.
Volume of water ($V_w$) = 1 liter = 1000 cm$^3$.
Use $\pi = 3.14$.

Since the vessel is vertex downwards, the water forms a smaller cone similar to the vessel.
The ratio of radius to height is constant: $\frac{r_w}{h_w} = \frac{R}{H}$.

The volume of the water cone is $V_w = \frac{1}{3}\pi r_w^2 h_w$.
Substitute $r_w = \frac{1}{2} h_w$:

Multiply both sides by 12:

Divide by 3.14:

Take the cube root:

The level of the water will be approximately $\boxed{\text{15.63 cm}}$.