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Let's go through each problem step by step.

### Problem 1:
This involves calculating the volume and surface area of a cube with edge length \(2\) meters.
1.1 **Volume of the cube**:  
The formula for the volume of a cube is:  
\[
V = a^3
\]  
Where \(a\) is the edge length. For \(a = 2\) meters:
\[
V = 2^3 = 8 \, \text{m}^3
\]

1.2 **Surface area of the cube**:  
The formula for the surface area of a cube is:  
\[
A = 6a^2
\]  
For \(a = 2\) meters:
\[
A = 6 \times 2^2 = 6 \times 4 = 24 \, \text{m}^2
\]

### Problem 2:
Here we deal with a parallelepiped (rectangular box) and must first determine the missing dimensions.

2.1 **Finding dimensions \(a\) and \(b\)**:  
- From the diagram, the height is \(3\) cm, the width is \(2\) cm, and the depth is \(4\) cm.  
- Since the unfolded figure (planification) shows faces corresponding to the sides of the parallelepiped, the missing dimensions are:
  - \(a\), which corresponds to the height (3 cm),
  - \(b\), which corresponds to the width (2 cm).

2.2 **Volume and surface area of the parallelepiped**:
2.2.1 **Volume**:  
The volume \(V\) of a parallelepiped is calculated using the formula:  
\[
V = l \times w \times h
\]  
Where \(l = 4\) cm, \(w = 2\) cm, and \(h = 3\) cm:
\[
V = 4 \times 2 \times 3 = 24 \, \text{cm}^3
\]

2.2.2 **Surface area**:  
The surface area \(A\) is given by:
\[
A = 2(lw + lh + wh)
\]
Substituting the given values:
\[
A = 2(4 \times 2 + 4 \times 3 + 2 \times 3) = 2(8 + 12 + 6) = 2 \times 26 = 52 \, \text{cm}^2
\]

### Problem 3:
This involves a pizza box, which is modeled as a right triangular prism.

3.1 **Classification of the triangle**:  
The base of the prism is a right triangle with one leg of \(21\) cm and the hypotenuse of \(28\) cm. We can check whether this triangle is a right triangle using the Pythagorean theorem:
\[
a^2 + b^2 = c^2
\]
Substituting \(a = 21\), \(b = 26\), and \(c = 28\):
\[
21^2 + 26^2 = 28^2 \quad \Rightarrow \quad 441 + 676 = 784 \quad \Rightarrow \quad 1117 \neq 784
\]
So, there seems to be an error with the height mentioned. Based on this, the triangle does not match the expected right triangle. For now, let's assume the triangle is not classified as such without further details.

3.2 **Volume of the pizza box**:  
The volume \(V\) of a triangular prism is given by:
\[
V = \text{Base area} \times \text{Height of the prism}
\]
The base area \(A_b\) of the triangle is:
\[
A_b = \frac{1}{2} \times \text{base} \times \text{height}
\]
Substituting the values:
\[
A_b = \frac{1}{2} \times 21 \times 26 = 273 \, \text{cm}^2
\]
The height of the prism is \(4.5\) cm, so the volume is:
\[
V = 273 \times 4.5 = 1228.5 \, \text{cm}^3
\]

---

Would you like more detailed explanations on any of these steps or have any further questions?

Here are 5 related questions to deepen your understanding:
1. How would the volume of the cube change if the edge length doubled?
2. Can you derive the surface area of the cube using an alternate method?
3. How would you calculate the diagonal of the parallelepiped in Problem 2?
4. How can we use the Pythagorean theorem to classify other triangles?
5. How would the volume of the pizza box change if its height doubled?

### Tip: 
Always double-check given dimensions and units to ensure accuracy, especially in geometric problems.

1. Calculate the volume and surface area of a cube with an edge length of 2 meters. 2. Determine the missing dimensions and find the volume and surface area of a parallelepiped given its planification. 3. Classify the triangle in a pizza box prism and calculate the volume of the box.

This guide explains how to calculate the volume and surface area of geometric shapes such as a cube, parallelepiped, and triangular prism. It includes step-by-step solutions and key formulas for students in Grades 8-10.

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