The diagonals of a convex quadrilateral $ABCD$ intersect at a point $O$. Find all angles of this quadrilateral if $\measuredangle OBA=30^{\circ},\measuredangle OCB=45^{\circ},\measuredangle ODC=45^{\circ}$, and $\measuredangle OAD=30^{\circ}$.
Given the angles $\measuredangle OBA = 30^\circ$, $\measuredangle OCB = 45^\circ$, $\measuredangle ODC = 45^\circ$, and $\measuredangle OAD = 30^\circ$, we need to find all the angles of the convex quadrilateral $ABCD$.

### Method 1:
1. **Using the Law of Sines in $\triangle BOC$ and $\triangle AOD$:**
   - In $\triangle BOC$, we have:
     \[
     \frac{BC}{\sin \alpha} = \frac{BO}{\sin 45^\circ}
     \]
   - In $\triangle AOD$, we have:
     \[
     \frac{AD}{\sin \alpha} = \frac{DO}{\sin 30^\circ}
     \]

2. **Dividing the two equations:**
   \[
   \frac{\frac{BC}{\sin \alpha}}{\frac{AD}{\sin \alpha}} = \frac{\frac{BO}{\sin 45^\circ}}{\frac{DO}{\sin 30^\circ}}
   \]
   Simplifying, we get:
   \[
   \frac{BC}{AD} = \frac{BO \cdot \sin 30^\circ}{DO \cdot \sin 45^\circ}
   \]
   Since $\sin 30^\circ = \frac{1}{2}$ and $\sin 45^\circ = \frac{\sqrt{2}}{2}$, we have:
   \[
   \frac{BC}{AD} = \frac{BO \cdot \frac{1}{2}}{DO \cdot \frac{\sqrt{2}}{2}} = \frac{BO}{DO \cdot \sqrt{2}}
   \]

3. **Using the given condition $\left( \frac{BC}{AD} \right)^2 = \frac{BO}{DO}$:**
   \[
   \left( \frac{BO}{DO \cdot \sqrt{2}} \right)^2 = \frac{BO}{DO}
   \]
   Simplifying, we get:
   \[
   \frac{BO^2}{2DO^2} = \frac{BO}{DO}
   \]
   Multiplying both sides by $2DO^2$, we get:
   \[
   BO^2 = 2BO \cdot DO
   \]
   Dividing both sides by $BO$, we get:
   \[
   BO = 2DO
   \]

4. **Returning to the previous relations:**
   - If $BO = 2DO$, then substituting back, we get:
     \[
     \frac{BC}{\sin \alpha} = \frac{2DO}{\sin 45^\circ}
     \]
     \[
     \frac{AD}{\sin \alpha} = \frac{DO}{\sin 30^\circ}
     \]
     Dividing these, we get:
     \[
     \frac{BC}{AD} = \frac{2DO \cdot \sin 30^\circ}{DO \cdot \sin 45^\circ} = \frac{2 \cdot \frac{1}{2}}{\frac{\sqrt{2}}{2}} = \frac{1}{\frac{\sqrt{2}}{2}} = \sqrt{2}
     \]

5. **Finding angle $\alpha$:**
   - From the above, we get:
     \[
     \sin \alpha = \sin 60^\circ \quad \text{or} \quad \sin \alpha = \sin 120^\circ
     \]
     Therefore, $\alpha = 60^\circ$ or $\alpha = 120^\circ$.

6. **Calculating the angles of the quadrilateral:**
   - If $\alpha = 60^\circ$:
     \[
     \angle A = 60^\circ, \quad \angle B = 180^\circ - 30^\circ - 45^\circ = 105^\circ, \quad \angle C = 60^\circ, \quad \angle D = 180^\circ - 30^\circ - 45^\circ = 135^\circ
     \]
   - If $\alpha = 120^\circ:
     \[
     \angle A = 120^\circ, \quad \angle B = 180^\circ - 30^\circ - 45^\circ = 45^\circ, \quad \angle C = 120^\circ, \quad \angle D = 180^\circ - 30^\circ - 45^\circ = 75^\circ
     \]

### Method 2:
1. **Observing the symmetry:**
   - Since $\measuredangle OBA = \measuredangle OAD = 30^\circ$ and $\measuredangle OCB = \measuredangle ODC = 45^\circ$, we can see that $\angle A = \angle C$ and $\angle B = \angle D$.

2. **Using the sum of angles in a quadrilateral:**
   - The sum of the angles in a quadrilateral is $360^\circ$.
   - Therefore, we have:
     \[
     2\angle A + 2\angle B = 360^\circ
     \]
     Simplifying, we get:
     \[
     \angle A + \angle B = 180^\circ
     \]

3. **Finding the angles:**
   - If $\angle A = 60^\circ$, then $\angle B = 120^\circ$.
   - If $\angle A = 120^\circ$, then $\angle B = 60^\circ$.

The final answer is $\boxed{\angle A = 60^\circ, \angle B = 105^\circ, \angle C = 60^\circ, \angle D = 135^\circ}$ or $\boxed{\angle A = 120^\circ, \angle B = 45^\circ, \angle C = 120^\circ, \angle D = 75^\circ}$.