Answer: 1.2
Solution: Draw $A O, P Q$ and $O Q$. We note that $A O \perp P Q$ since $A Q=A P$ and $O Q=O P$.
This implies that $\angle P A O \cong \angle E P Q$ (both complementary to $\angle A P R$ )
Hence $\triangle P A O \approx \triangle E P Q$. Therefore
$$
\frac{A P}{O P}=\frac{P E}{Q E} \Longrightarrow Q E=\frac{P E \times O P}{A P} \Longrightarrow Q E=\frac{3.6 \times 2}{6}=1.2
$$
3. (Easy/Average) The segments $A P$ and $A Q$ are tangent to circle $O$ at points $P$ and $Q$, respectively. Moreover, $Q E$ is perpendicular to diameter $P D$ of length 4 . If $P E=3.6$ and $A P=6$, what is the length of $Q E$ ?