7.54 Suppose the radius of the small circle is $\frac{r}{2}$, and the radius of the large circle is $r$. How many small circles are needed at minimum to completely cover the area of the large circle?
[Solution] (1) A circular surface with radius $r$ can be covered by 7 small circular discs with radius $\frac{r}{2}$. Construct a regular hexagon inscribed in the large circle $O$, and then, with the midpoints of each side of the hexagon and point $O$ as centers, and $\frac{r}{2}$ as the radius, draw 7 small circles. These will cover the entire large circular surface (see the figure below).

In fact, connect $O A, O B, O C$. Let the midpoints of $O B$ and $O C$ be $M$ and $L$, respectively. Then $O B = O C = r$, $O M = M B = O L = L C = \frac{r}{2}$. Connect $H M, M G$, then $\triangle B H M$ and $\triangle B M G$ are both equilateral triangles, so $H M = M G = \frac{r}{2}$, which means point $M$ is exactly the intersection of circles $H$, $G$, and the small circle $O$. Similarly, point $L$ is the intersection of circles $H$, $I$, and the small circle $O$. It is evident that the sector $O B C$ is covered by circles $H$ and $O$. Therefore, the entire large circular surface is completely covered by 7 small circular discs.
(2) Below is the proof that any 6 small circular discs with radius $\frac{r}{2}$ cannot completely cover a large circular surface with radius $r$.

Divide the circumference of the large circle into 48 equal parts, obtaining 48 points. Consider these 48 points and the center $O$, making a total of 49 points. Clearly, when two points are separated by 8 arcs, the distance between the two points is $r$. Therefore, a small circular disc with radius $\frac{r}{2}$ can cover at most 9 of the 48 points. Additionally, the distance from the center $O$ to any point is $r$, so a small circular disc covering point $O$ can cover at most 1 point. Therefore, no matter how the 6 small circular discs with radius $\frac{r}{2}$ are placed, they cannot cover all 49 points.
In summary, at least 7 small circular discs are required to completely cover the large circular surface.