(i) Find the angles of $\triangle ABC$ if the length of the altitude through $B$ is equal to the length of the median through $C$ and the length of the altitude through $C$ is equal to the length of the median through $B$.

(ii) Find all possible values of $\angle ABC$ of $\triangle ABC$ if the length of the altitude through $A$ is equal to the length of the median through $C$ and the length of the altitude through $C$ is equal to the length of the median through $B$.
### Part (i)

Given:
- The length of the altitude through \( B \) is equal to the length of the median through \( C \).
- The length of the altitude through \( C \) is equal to the length of the median through \( B \).

We need to find the angles of \(\triangle ABC\).

1. **Define the sides and medians:**
   Let \( AB = c \), \( AC = b \), and \( BC = a \).

2. **Express the altitude and median:**
   - The altitude from \( B \) to \( AC \) is \( h_B \).
   - The median from \( C \) to \( AB \) is \( m_C \).
   - The altitude from \( C \) to \( AB \) is \( h_C \).
   - The median from \( B \) to \( AC \) is \( m_B \).

3. **Equate the given conditions:**
   \[
   h_B = m_C \quad \text{and} \quad h_C = m_B
   \]

4. **Formulas for altitudes and medians:**
   - Altitude \( h_B \):
     \[
     h_B = \frac{2}{a} \sqrt{s(s-a)(s-b)(s-c)}
     \]
     where \( s = \frac{a+b+c}{2} \) is the semi-perimeter.
   - Median \( m_C \):
     \[
     m_C = \frac{1}{2} \sqrt{2a^2 + 2b^2 - c^2}
     \]

5. **Equate \( h_B \) and \( m_C \):**
   \[
   \frac{2}{a} \sqrt{s(s-a)(s-b)(s-c)} = \frac{1}{2} \sqrt{2a^2 + 2b^2 - c^2}
   \]

6. **Simplify the equation:**
   \[
   4 \sqrt{s(s-a)(s-b)(s-c)} = a \sqrt{2a^2 + 2b^2 - c^2}
   \]

7. **Similarly, equate \( h_C \) and \( m_B \):**
   \[
   \frac{2}{a} \sqrt{s(s-a)(s-b)(s-c)} = \frac{1}{2} \sqrt{2a^2 + 2c^2 - b^2}
   \]

8. **Simplify the equation:**
   \[
   4 \sqrt{s(s-a)(s-b)(s-c)} = a \sqrt{2a^2 + 2c^2 - b^2}
   \]

9. **Conclusion:**
   Since the conditions are symmetric and the only way for the altitudes and medians to be equal is if the triangle is equilateral, we conclude:
   \[
   AB = BC = CA
   \]
   Therefore, all angles must be \(60^\circ\).

### Part (ii)

Given:
- The length of the altitude through \( A \) is equal to the length of the median through \( C \).
- The length of the altitude through \( C \) is equal to the length of the median through \( B \).

We need to find all possible values of \(\angle ABC\).

1. **Define the sides and medians:**
   Let \( AB = c \), \( AC = b \), and \( BC = a \).

2. **Express the altitude and median:**
   - The altitude from \( A \) to \( BC \) is \( h_A \).
   - The median from \( C \) to \( AB \) is \( m_C \).
   - The altitude from \( C \) to \( AB \) is \( h_C \).
   - The median from \( B \) to \( AC \) is \( m_B \).

3. **Equate the given conditions:**
   \[
   h_A = m_C \quad \text{and} \quad h_C = m_B
   \]

4. **Formulas for altitudes and medians:**
   - Altitude \( h_A \):
     \[
     h_A = \frac{2}{a} \sqrt{s(s-a)(s-b)(s-c)}
     \]
   - Median \( m_C \):
     \[
     m_C = \frac{1}{2} \sqrt{2a^2 + 2b^2 - c^2}
     \]

5. **Equate \( h_A \) and \( m_C \):**
   \[
   \frac{2}{a} \sqrt{s(s-a)(s-b)(s-c)} = \frac{1}{2} \sqrt{2a^2 + 2b^2 - c^2}
   \]

6. **Simplify the equation:**
   \[
   4 \sqrt{s(s-a)(s-b)(s-c)} = a \sqrt{2a^2 + 2b^2 - c^2}
   \]

7. **Similarly, equate \( h_C \) and \( m_B \):**
   \[
   \frac{2}{a} \sqrt{s(s-a)(s-b)(s-c)} = \frac{1}{2} \sqrt{2a^2 + 2c^2 - b^2}
   \]

8. **Simplify the equation:**
   \[
   4 \sqrt{s(s-a)(s-b)(s-c)} = a \sqrt{2a^2 + 2c^2 - b^2}
   \]

9. **Conclusion:**
   By solving these equations, we find that the possible values for \(\angle ABC\) are \(60^\circ\) and \(120^\circ\).

The final answer is \( \boxed{ 60^\circ } \) and \(120^\circ\).