Let $A_1A_2A_3$ be an acute-angled triangle inscribed into a unit circle centered at $O$. The cevians from $A_i$ passing through $O$ meet the opposite sides at points $B_i$ $(i = 1, 2, 3)$ respectively.
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[*] Find the minimal possible length of the longest of three segments $B_iO$.
[*] Find the maximal possible length of the shortest of three segments $B_iO$.
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1. **Given**: \( A_1A_2A_3 \) is an acute-angled triangle inscribed in a unit circle centered at \( O \). The cevians from \( A_i \) passing through \( O \) meet the opposite sides at points \( B_i \) \((i = 1, 2, 3)\) respectively.
2. **Objective**: 
   - (a) Find the minimal possible length of the longest of three segments \( B_iO \).
   - (b) Find the maximal possible length of the shortest of three segments \( B_iO \).

### Part (a): Minimal possible length of the longest of three segments \( B_iO \)

3. **Using Gergonne-Euler Theorem**: 
   \[
   \frac{OB_1}{1 + OB_1} + \frac{OB_2}{1 + OB_2} + \frac{OB_3}{1 + OB_3} = 1
   \]
   This can be rewritten as:
   \[
   \sum_{i=1}^{3} \left( 1 - \frac{1}{1 + OB_i} \right) = 1 \implies \sum_{i=1}^3 \frac{1}{1 + OB_i} = 2
   \]

4. **Analyzing the inequality**:
   \[
   2 = \sum_{i=1}^3 \frac{1}{1 + OB_i} \geq \frac{3}{1 + OB_1}
   \]
   This implies:
   \[
   1 + OB_1 \geq \frac{3}{2} \implies OB_1 \geq \frac{1}{2}
   \]

5. **Conclusion for part (a)**:
   The minimal possible length of the longest segment \( OB_1 \) is \( \boxed{\frac{1}{2}} \).

### Part (b): Maximal possible length of the shortest of three segments \( B_iO \)

6. **Analyzing the inequality**:
   \[
   2 = \sum_{i=1}^3 \frac{1}{1 + OB_i} \leq \frac{3}{1 + OB_3}
   \]
   This implies:
   \[
   1 + OB_3 \leq \frac{3}{2} \implies OB_3 \leq \frac{1}{2}
   \]

7. **Conclusion for part (b)**:
   The maximal possible length of the shortest segment \( OB_3 \) is \( \boxed{\frac{1}{2}} \).

8. **Verification**:
   It can be easily seen that one case of equality is when \( \triangle A_1A_2A_3 \) is equilateral. In this case, all \( OB_i \) are equal and each \( OB_i = \frac{1}{2} \).