1. Instead of truncated pyramids, we examine the pyramids defined by the apex of the original pyramid and planes parallel to its base. Let the distances of these planes from the common apex be $m_{1}, m_{2}, \ldots, m_{24} \left( m_{24} = m \right.$, the height of the original pyramid). If the volume of the original pyramid is $V$, then the volume of the pyramid at height $m_{i}$ according to the division is $V_{i} = \frac{i}{24} V$, so $\frac{V_{i}}{V} = \frac{i}{24}, i = 1, 2, \ldots, 23$.

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The triangles forming the sides of the pyramids and the base triangles are similar, with the ratio of corresponding edges being $\frac{m_{i}}{m}$, hence $\frac{V_{i}}{V} = \frac{m_{i}^{3}}{m^{3}}$. From the two equalities, it follows that

$$m_{i} = m \sqrt[3]{\frac{i}{24}}$$

In the second division, the lateral surface area of the pyramid at height $m_{k}^{\prime}$ is $F_{k} = \frac{k}{24} F$, where $F$ is the lateral surface area of the original pyramid, $k = 1, 2, \ldots, 23$, so $\frac{F_{k}}{F} = \frac{k}{24}$. By similarity, $\frac{F_{k}}{F} = \frac{m_{k}^{\prime 2}}{m^{2}}$, hence $m_{k}^{\prime} = m \sqrt{\frac{k}{24}}$.

The two sets of planes have a common plane if for some pair of numbers $i, k$ it holds that $m_{i} = m_{k}^{\prime}$. This means that

$$\sqrt[3]{\frac{i}{24}} = \sqrt{\frac{k}{24}}, \quad \text{or} \quad 24 i^{2} = k^{3}$$

Since $24 = 3 \cdot 2^{3}$, the pair $i = 3, k = 6$ satisfies the condition: the third plane of the first division coincides with the sixth plane of the second division. This plane bisects the pyramid at half its height, above it is the $\left(\frac{1}{2}\right)^{3} = \frac{3}{24}$ part of the volume $V$ and the $\left(\frac{1}{2}\right)^{2} = \frac{6}{24}$ part of the lateral surface area $F$.

2. If the pyramid is divided into parts by $n$ planes according to the two principles, then similarly to the above, the condition is the equation

$$(n+1) i^{2} = k^{3}, \quad 1 \leq i, \quad k \leq n$$

For which values of $(n+1)$ does the two $n$-element sets of planes have a common plane?

If $(n+1)$ in its prime factorization does not contain a third or higher power, then the equation has no solution under the condition $1 \leq i, k \leq n$. In this case, in the factorization of $(n+1)$, the exponent of every prime number is either $\alpha = 1$ or $\alpha = 2$, and $i^{2}$ must (for the purpose of $k^{3}$) raise it to 3 or a multiple of 3, from $\alpha = 1$ to at least 3, and from $\alpha = 2$ to at least 6, since the exponent in $i^{2}$ is even. Thus, $i$ must contain every prime factor of $(n+1)$ with at least as large an exponent as $(n+1)$ itself, which is only possible if $i \geq n+1$.

However, if $n+1 = q \cdot p^{3}$, where $p > 1$ and $q$ are natural numbers, then for example, the pair $i = q, k = q \cdot p$ is a solution to the equation for every $q = 1, 2, 3, \ldots$. The cubes less than 100 are $8, 27, 64$, so the corresponding values of $n$ are:

$7, 15, 23, 26, 31, 39, 47, 53, 55, 63, 71, 79, 80, 87, 95$.

We already notice here that $n = 63$ can be obtained in two different ways: $63 = 8 \cdot 2^{3} - 1 = 1 \cdot 4^{3} - 1$, so here we get two common planes: $i = 8, k = 16$ and $i = 1, k = 4$.