1. Identify the problem: We need to determine if there exists a point on the plane such that the distances from this point to the vertices of a square are exactly 1, 4, 7, and 8 units.

2. Use the distance formula: Let the vertices of the square be $A, B, C,$ and $D$. Let the point be $P$. The distances from $P$ to $A, B, C,$ and $D$ are given as 1, 4, 7, and 8 units.

3. Apply the distance relationship: For a point $P$ to have these distances to the vertices of a square, the sum of the squares of the distances from $P$ to opposite vertices must be equal. This is derived from the properties of a rectangle (and hence a square).

4. Check the sum of squares:
$$1^2 + 8^2 = 1 + 64 = 65$$
$$4^2 + 7^2 = 16 + 49 = 65$$
Both sums are equal, which is a necessary condition for the distances to be from a point to the vertices of a rectangle.

5. Analyze the side length of the square: For a square, the side length $s$ must satisfy the Pythagorean theorem in the context of the distances from $P$ to the vertices. The side length $s$ must be such that the distances from $P$ to the vertices form a valid geometric configuration.

6. Determine the range of side lengths: The side length $s$ of the square must be in the range where the distances from $P$ to the vertices can form a square. This involves checking the possible configurations:
- For a square with side length $s$, the distances from $P$ to the vertices must fit within the constraints of the square's geometry.
- The side length $s$ must be such that the distances 1, 4, 7, and 8 can be realized.

7. Check the feasibility: Given the distances 1, 4, 7, and 8, we need to check if there exists a side length $s$ that fits within the constraints:
- The side length $s$ must be in the range of $(3, 5)$ and $(6, 8)$ to satisfy the distance conditions.
- However, no single side length $s$ can satisfy both ranges simultaneously.

Conclusion:
Since the side length of the square must fit within both ranges simultaneously, and this is impossible, the given distances cannot be from a point to the vertices of a square.

The final answer is False.