Let $ABC$ be an acute-angled triangle with $AB<AC$. Let $D$ be the midpoint of side $BC$ and $BE,CZ$ be the altitudes of the triangle $ABC$. Line $ZE$ intersects line $BC$ at point $O$.
(i) Find all the angles of the triangle  $ZDE$ in terms of angle  $\angle A$ of the triangle $ABC$
(ii) Find the angle $\angle BOZ$ in terms of angles   $\angle B$ and   $\angle C$ of the triangle $ABC$
Let's solve the problem step-by-step.

### Part (i): Find all the angles of the triangle $ZDE$ in terms of angle $\angle A$ of the triangle $ABC$.

1. **Identify the concyclic points:**
   Since $\angle BEC = \angle BZC = 90^\circ$, points $B, C, E, Z$ are concyclic. This means that $B, C, E, Z$ lie on a circle with $D$ as the center (since $D$ is the midpoint of $BC$).

2. **Isosceles triangles:**
   Because $D$ is the center of the circle passing through $B, C, E, Z$, triangles $BDZ$, $ZDE$, and $EDC$ are isosceles with $BD = DZ$, $ZD = DE$, and $DE = DC$ respectively.

3. **Angle chasing:**
   - $\angle BDC = 180^\circ - 2\angle B$ (since $D$ is the midpoint and $BD = DC$).
   - $\angle BZD = \angle BZD = 180^\circ - 2\angle B$ (since $BDZ$ is isosceles).
   - $\angle CDE = 180^\circ - 2\angle C$ (since $EDC$ is isosceles).

4. **Calculate $\angle EZD$:**
   \[
   \angle EZD = 180^\circ - \angle BDZ - \angle CDE = 180^\circ - (180^\circ - 2\angle B) - (180^\circ - 2\angle C)
   \]
   Simplifying,
   \[
   \angle EZD = 180^\circ - 180^\circ + 2\angle B - 180^\circ + 2\angle C = 2\angle B + 2\angle C - 180^\circ
   \]
   Since $\angle A + \angle B + \angle C = 180^\circ$,
   \[
   \angle EZD = 2(180^\circ - \angle A) - 180^\circ = 360^\circ - 2\angle A - 180^\circ = 180^\circ - 2\angle A
   \]

5. **Calculate $\angle ZED$:**
   Since $\triangle ZDE$ is isosceles with $ZD = DE$,
   \[
   \angle ZED = \angle EZD = 180^\circ - 2\angle A
   \]

6. **Calculate $\angle DZE$:**
   Using the fact that the sum of angles in a triangle is $180^\circ$,
   \[
   \angle DZE = 180^\circ - \angle EZD - \angle ZED = 180^\circ - (180^\circ - 2\angle A) - (180^\circ - 2\angle A) = 2\angle A
   \]

### Part (ii): Find the angle $\angle BOZ$ in terms of angles $\angle B$ and $\angle C$ of the triangle $ABC$.

1. **Identify the angles:**
   Since $O$ is the intersection of $ZE$ and $BC$, and $ZE$ is the radical axis of the circle passing through $B, C, E, Z$,
   \[
   \angle BOZ = \angle COE
   \]

2. **Angle chasing:**
   \[
   \angle COE = 180^\circ - \angle CEO - \angle C
   \]
   Since $\angle CEO = \angle ZED$ and $\angle ZED = 180^\circ - 2\angle A$,
   \[
   \angle COE = 180^\circ - (180^\circ - 2\angle A) - \angle C = 2\angle A - \angle C
   \]

3. **Simplify:**
   Using $\angle A + \angle B + \angle C = 180^\circ$,
   \[
   \angle BOZ = 2\angle A - \angle C = 2(180^\circ - \angle B - \angle C) - \angle C = 360^\circ - 2\angle B - 2\angle C - \angle C = 360^\circ - 2\angle B - 3\angle C
   \]

The final answer is $\boxed{2\angle A - \angle C}$