4. Let's introduce the notations: $B H=2 a, H C=a, B F=y, F C=x$. Since angle $B F H$ is a right angle, by the theorem of relations in a right-angled triangle for the two legs $B H, H C$, we have:

$$
\left\{\begin{array}{l}
a^{2}=x(y+x) \\
4 a^{2}=y(x+y)
\end{array} \quad \Rightarrow \frac{y}{x}=4 \Rightarrow y=4 x\right.
$$

From the ratio of the areas of triangles with a common angle, we find the answer to the first question:

$$
\frac{S_{F C H}}{S_{A B C}}=\frac{x}{2(x+y)}=\frac{1}{10}
$$

By the Pythagorean theorem for triangle $B H C$, we find $x: 5 a^{2}=25 x^{2} \Rightarrow x=$ $\frac{a}{\sqrt{5}}$. Let $O$ be the center of the circle circumscribed around $\triangle B H F$. Denote $\delta=\angle H O F$. Then, by the cosine theorem for triangle $B O F$:

$$
\left(\frac{4 a}{\sqrt{5}}\right)^{2}=2 a^{2}+2 a^{2} \cos \delta \Rightarrow \cos \delta=\frac{3}{5} \Rightarrow \sin \delta=\frac{4}{5}
$$

Denote the area of sector $H O F$ by $S_{1}$. Then $S_{1}=\frac{1}{2} a^{2} \cdot \arcsin \frac{4}{5}$. Inside the circle, there are two such sectors for the triangle. In addition, inside the circle, there are two triangles of equal area. Let's find the area $S_{2}$ of triangle $B O F: S_{2}=\frac{1}{2} a^{2} \sin \delta=\frac{2}{5} a^{2}$. Then the answer to the second question will be: $\frac{2\left(S_{1}+S_{2}\right)}{S_{A B C}}=\frac{1}{2}\left(\arcsin \frac{4}{5}+\frac{4}{5}\right)$.
Answer: $\frac{1}{10} ; \frac{2\left(S_{1}+S_{2}\right)}{S_{A B C}}=\frac{1}{2}\left(\arcsin \frac{4}{5}+\frac{4}{5}\right)$. Answer to the variant: $4-2: \frac{1}{20}$; $\frac{3}{4}\left(\arcsin \frac{3}{5}+\frac{3}{5}\right)$