The first thing I want to say is whether there is a 25th-order 4-regular planar graph. I found some websites on the Internet, trying to find some clues. Indeed, I found some.
I think the second website is great！
In the second URL, surprisingly, there is no any relevant information about the 4-regular planar graph with 25 vertices.
I also try to solve it programmatically, but it seems relatively difficult, the number of vertices is a bit large. I tried to derive the contradiction through Euler’s formula, but failed.
plantri
If it does not exist, then the following question seems to become meaningless.
Actually we want to find a simple $4$-regular plane graph with $25$ vertices, which satisfies the following conditions:
(1) Each face is either a 3-degree face or a 4-degree face.
(2) There is exactly one vertex associated with all 4-degree faces, and all the other vertices are associated with ** one 3-degree face** and three 4-degree face.
Of course it may not exist.
PS: We can derive some information about the number of faces
Combine the following two formulas
$$3 f_3+4 f_4=2m=25\times 4$$
$$25- \frac{4\times 25}{2}+f_3+f_4=2$$
We get the number of three-degree face is $f_3=8$ and the number of three-degree face is $f_4=19$.
Best Answer
[Edit:] A previous version of this question asked about graphs where the 24 nonspecial vertices were each on 3 $3$-degree faces and a single $4$-degree face.
This is impossible.
Let $\Gamma$ be such a graph, and let $u$ be the unique vertex that is on four $4$-degree faces. A simple counting argument shows that $\Gamma$ has 24 $3$-degree faces and $7$ $4$-degree faces. This means that there is at least one $4$-degree face $F$ that does not contain $u$. But the fact that each vertex on $F$ is only on other $3$-degree faces forces the structure of this graph, there will be only 8 vertices.
(I don't have the graph drawing ready, it will be a four-cycle with triangles on each edge, there will be one way to draw the remaining four edges.)