A) The height of the pyramid from which the frustum was cut from. b) The surface area of the frustum

Let $H$ be the height of the original large pyramid and $h_s$ be the height of the small pyramid that was cut off.
The height of the frustum is $h = 60$ cm.
The side length of the bottom base of the frustum (and the large pyramid) is $L_1 = 40$ cm.
The side length of the top base of the frustum (and the small pyramid) is $L_2 = 20$ cm.
The height of the large pyramid is $H = h_s + h$.

Consider a cross-section of the pyramid. By similar triangles, the ratio of the heights to the base side lengths (or half-side lengths) is constant:

Substitute $h_s = H - h$:

Substitute the given values $L_1 = 40$ cm, $L_2 = 20$ cm, and $h = 60$ cm:

Multiply both sides by 40:

The height of the pyramid from which the frustum was cut is $\boxed{\text{120 cm}}$.

The frustum is an open container, meaning it has no top base. Its surface area consists of the area of the bottom base and the area of the four trapezoidal lateral faces.
Area of bottom base ($A_B$):

Consider a right triangle formed by the height of the frustum ($h$), the difference in half-base lengths, and the slant height of the frustum ($l_f$).
The horizontal leg of this triangle is $\frac{L_1 - L_2}{2} = \frac{40 - 20}{2} = \frac{20}{2} = 10$ cm.
Using the Pythagorean theorem:

The area of a trapezoid is given by $\frac{1}{2} (b_1 + b_2) h'$, where $b_1$ and $b_2$ are the parallel sides and $h'$ is the height (which is the slant height $l_f$ in this case).

There are 4 identical lateral faces:

The surface area of the frustum is $\boxed{1600 + 1200\sqrt{37} \text{ cm}^2}$.