1. Solution. First, consider some fixed vertical order of laying the glasses (from bottom to top). It is clear that by rotating the entire layout by some angle, we will not change the layout (we will not get a new layout). Therefore, we can assume that the bottom glass in the layout is always fixed (does not rotate). Then different layouts (with a fixed vertical order of glasses) will differ from each other by the rotation of each of the subsequent 4 upper glasses by $0^{\circ}$, $120^{\circ}$, and $240^{\circ}$ relative to the fixed bottom glass. Therefore, we will get a total of $3^{4}=81$ layout options with a fixed vertical order of glasses. Taking into account all possible vertical permutations of the five glasses (5! = 120 options), we get the total number of possible glass layouts in a stack: 5! $3^{4}=120 \cdot 81=9720$ pieces. But not all these layouts meet the condition of the problem. The condition of the problem is satisfied by all layouts except those in which the entire stack turns out to be completely vertically opaque. Let's first find the number of such layouts.

For this, consider columns of isosceles triangles (parts located above each of the three fixed isosceles triangles of the bottom glass). Let's determine the number of such layouts in which each of these four columns is opaque. One of these columns is already opaque by default (the one standing on the bottom shaded isosceles triangle). Consider the ordered set (vector) of possible rotation angles (for definiteness, clockwise) of the shaded isosceles triangles on each of the 4 upper glasses relative to the bottom shaded equilateral triangle: $\left(\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}\right)$, where $\alpha_{k} \in\left\{0^{\circ}, 120^{\circ}, 240^{\circ}\right\}$, $k=1,2,3,4$.

All "columns" of glass isosceles triangles will be opaque if this set of rotation angles necessarily includes at least one $120^{\circ}$ and at least one $240^{\circ}$ (while the angle $0^{\circ}$ is not required to appear, although it will not hurt if it does appear). The other two places in the set can be occupied by any of the three possible values $\left\{0^{\circ}, 120^{\circ}, 240^{\circ}\right\}$, i.e., $0^{\circ}$ and $120^{\circ}$, or $0^{\circ}$ and $240^{\circ}$, or $120^{\circ}$ and $240^{\circ}$, as well as $0^{\circ}$ and $0^{\circ}$, or $120^{\circ}$ and $120^{\circ}$, or $240^{\circ}$ and $240^{\circ}$. Therefore, to count the total number of such sets (and thus the number of layouts with a fixed vertical order of glasses), we will divide them into the following six groups:

- If the two additional values are $0^{\circ}$ and $120^{\circ}$, then we get sets in which the rotation angles $0^{\circ}$ and $240^{\circ}$ appear once, and the angle $120^{\circ}$ appears twice. The total number of such sets (due to permutations) will be $6 \cdot 2=12$ (here 6 is the number of ways to choose places in the vector for two angles of $120^{\circ}$, and 2 is the number of ways to permute the angles $0^{\circ}$ and $240^{\circ}$ in the two remaining places);
- If the two additional values are $0^{\circ}$ and $240^{\circ}$, then we get sets in which the rotation angles $0^{\circ}$ and $120^{\circ}$ appear once, and the angle $240^{\circ}$ appears twice. Analogously to the previous case, the total number of such sets (due to permutations) will be $6 \cdot 2=12$;
- If the two additional values are $120^{\circ}$ and $240^{\circ}$, then we get sets in which the rotation angles $120^{\circ}$ and $240^{\circ}$ appear twice each. The total number of such sets (due to permutations) will be $6 \cdot 1=6$ (here 6 is the number of ways to choose two places out of four for placing two angles of $120^{\circ}$, and 1 is the only way to place two angles of $240^{\circ}$ in the two remaining places);
- If the two additional values are $0^{\circ}$ and $0^{\circ}$, then we get sets in which the rotation angles $120^{\circ}$ and $240^{\circ}$ appear once, and the angle $0^{\circ}$ appears twice. The total number of such sets (due to permutations) will be $6 \cdot 2=12$ (here 6 is the number of ways to choose two places out of four for placing two angles of $0^{\circ}$, and 2 is the number of ways to permute the angles $120^{\circ}$ and $240^{\circ}$ in the two remaining places);
- If the two additional values are $120^{\circ}$ and $120^{\circ}$, then we get sets in which the rotation angle $240^{\circ}$ appears once, and the angle $120^{\circ}$ appears three times. The total number of such sets will be $4 \cdot 1=4$ (here 4 is the number of ways to choose one place out of four for placing the angle $240^{\circ}$, and 1 is the only way to place the identical angles $120^{\circ}$ in the three remaining places);
- If the two additional values are $240^{\circ}$ and $240^{\circ}$, then we get sets in which the rotation angle $120^{\circ}$ appears once, and the angle $240^{\circ}$ appears three times. Analogously to the previous case, the total number of such sets will be $4 \cdot 1=4$.

In the end, we get a total of $12+12+6+12+4+4=50$ ordered sets of rotation angles that satisfy the required condition. This is the total number of layouts with a fixed vertical order of glasses in which the final stack turns out to be opaque. By permuting the glasses (in the vertical direction) in $5!=120$ ways, we get the total number of $120 \cdot 50=6000$ layouts in which the final stack turns out to be opaque. Finally, by subtracting the number of 6000 unwanted layouts from the total number of different layouts 9720, we finally get $9720-6000=3720$ ways to lay out a partially transparent stack of five glasses.
Answer: 3720 ways.

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