$$
P(E \cup F)-P(E \cap F)=\frac{11}{25}
$$
(i.e. only $E$ or only $F$ )

Neither of them occurs $=\frac{2}{25}$ $\Rightarrow \quad P(\bar{E} \cap \bar{F})=\frac{2}{25}$
From Eq. (i), we get
$$
P(E)+P(F)-2 P(E \cap F)=\frac{11}{25}
$$
From Eq. (ii), we get
$$
\begin{gathered}
(1-P(E))(1-P(F))=\frac{2}{25} \\
\Rightarrow 1-P(E)-P(F)+P(E) \cdot P(F)=\frac{2}{25}
\end{gathered}
$$
(iv)
From Eqs. (iii) and (iv), we get
$$
\begin{aligned}
& P(E)+P(F)=\frac{7}{5} \text { and } P(E) \cdot P(F)=\frac{12}{25} \\
& \therefore \quad P(E) \cdot\left\{\frac{7}{5}-P(E)\right\}=\frac{12}{25}
\end{aligned}
$$

$$
\begin{aligned}
& \Rightarrow \quad(P(E))^2-\frac{7}{5} P(E)+\frac{12}{25}=0 \\
& \Rightarrow \quad\left(P(E)-\frac{3}{5}\right)\left(P(E)-\frac{4}{5}\right)=0 \\
& \therefore \quad P(E)=\frac{3}{4} \text { or } \frac{4}{5} \Rightarrow P(F)=\frac{4}{5} \text { or } \frac{3}{5}
\end{aligned}
$$