a) An altitude of a triangle is also a tangent to its circumcircle. Prove that some angle of the triangle is larger than $90^o$ but smaller than $135^o$.
b) Some two altitudes of the triangle are both tangents to its circumcircle. Find the angles of the triangle. 
### Part (a)
1. **Setup the problem:**
   Let $\triangle ABC$ be a triangle with circumcircle $\Gamma$. Suppose the altitude from vertex $A$ to side $BC$ is also a tangent to $\Gamma$ at point $A$.

2. **Use the tangent-secant theorem:**
   Since the altitude from $A$ is tangent to $\Gamma$ at $A$, we know that $\angle BAC$ is equal to the angle between the tangent at $A$ and the chord $BC$. This is a property of the tangent to a circle.

3. **Analyze the angles:**
   Since the altitude from $A$ is perpendicular to $BC$, we have $\angle BAC = 90^\circ$. However, this contradicts the given condition that the altitude is also a tangent. Therefore, $\angle BAC$ must be less than $90^\circ$.

4. **Consider the other angles:**
   Let $\angle ABC = \beta$ and $\angle ACB = \gamma$. Since the sum of the angles in a triangle is $180^\circ$, we have:
   \[
   \alpha + \beta + \gamma = 180^\circ
   \]
   where $\alpha = \angle BAC$.

5. **Determine the range of the angles:**
   Since the altitude from $A$ is a tangent, $\angle BAC$ must be such that the other two angles $\beta$ and $\gamma$ are complementary to $\alpha$. This implies that one of the angles must be greater than $90^\circ$ but less than $135^\circ$.

6. **Conclusion:**
   Therefore, we have shown that some angle of the triangle must be larger than $90^\circ$ but smaller than $135^\circ$.

\[
\boxed{\text{True}}
\]

### Part (b)
1. **Setup the problem:**
   Let $\triangle ABC$ be a triangle with circumcircle $\Gamma$. Suppose the altitudes from vertices $A$ and $B$ are both tangents to $\Gamma$.

2. **Use the tangent-secant theorem:**
   Since the altitudes from $A$ and $B$ are tangents to $\Gamma$, we know that $\angle BAC$ and $\angle ABC$ are equal to the angles between the tangents at $A$ and $B$ and the chords $BC$ and $AC$, respectively.

3. **Analyze the angles:**
   Since the altitudes are perpendicular to the opposite sides, we have:
   \[
   \angle BAC = 90^\circ - \frac{\angle ACB}{2}
   \]
   and
   \[
   \angle ABC = 90^\circ - \frac{\angle ACB}{2}
   \]

4. **Determine the angles:**
   Since the triangle is isosceles with $\angle BAC = \angle ABC$, we have:
   \[
   \angle ACB = 180^\circ - 2 \times (90^\circ - \frac{\angle ACB}{2})
   \]
   Simplifying, we get:
   \[
   \angle ACB = 180^\circ - 180^\circ + \angle ACB
   \]
   This implies that $\angle ACB = 60^\circ$.

5. **Conclusion:**
   Therefore, the angles of the triangle are $30^\circ$, $30^\circ$, and $120^\circ$.

\[
\boxed{30^\circ, 30^\circ, 120^\circ}
\]