Let $ABC$ be a triangle with $\angle C=90^\circ$, and $A_0$, $B_0$, $C_0$ be the mid-points of sides $BC$, $CA$, $AB$ respectively. Two regular triangles $AB_0C_1$ and $BA_0C_2$ are constructed outside $ABC$. Find the angle $C_0C_1C_2$.
1. **Define the given angles and midpoints:**
   - Given $\angle C = 90^\circ$ in $\triangle ABC$.
   - $A_0$, $B_0$, and $C_0$ are the midpoints of $BC$, $CA$, and $AB$ respectively.
   - Construct regular triangles $AB_0C_1$ and $BA_0C_2$ outside $\triangle ABC$.

2. **Establish the right angles:**
   - $\angle B_0C_0A_0 = 90^\circ$ because $A_0$, $B_0$, and $C_0$ are midpoints, and $\angle C = 90^\circ$.
   - By Thales' theorem, $\angle C_0B_0C = \angle C_0A_0C = 90^\circ$.

3. **Analyze the regular triangles:**
   - $\angle CC_1A = 90^\circ$ since $C_1B_0 = B_0A = B_0C$.
   - Let $\omega_1$ be the circumcircle of $\triangle AC_1C$ and let $\omega_1 \cap BC = F$.
   - Since $AC$ is a diameter of $\omega_1$, $\angle AFC = 90^\circ$, making $F$ the foot of the altitude from $A$.

4. **Symmetric properties and congruence:**
   - Similarly, let $\omega_2$ be the circumcircle of $\triangle CC_2B$ and $F \in \omega_2$.
   - By Thales' theorem, $B_0C_0 = \frac{1}{2} BC = A_0C = A_0C_2$.
   - Similarly, $A_0C_0 = \frac{1}{2} AC = B_0A = B_0C_1$.

5. **Congruence of triangles:**
   - $\angle C_1B_0C_0 = \angle C_1B_0A + \angle AB_0C_0 = 60^\circ + 90^\circ = 150^\circ$.
   - $\angle C_0A_0B = \angle BA_0C_2 = \angle C_0A_0C_2 = 150^\circ$.
   - Thus, $\triangle C_1B_0C_0 \cong \triangle C_0A_0C_2$ by SAS congruence.

6. **Calculate the angle $C_0C_1C_2$:**
   - From the congruence, $C_0C_1 = C_0C_2$.
   - $\angle C_1C_0C_2 = \angle C_1C_0B_0 + \angle B_0C_0A_0 + \angle A_0C_0C_2 = 90^\circ + (\angle C_1C_0B_0 + \angle A_0C_0C_2) = 90^\circ + 30^\circ = 120^\circ$.
   - Therefore, $\angle C_0C_1C_2 = \frac{1}{2}(180^\circ - 120^\circ) = 30^\circ$.

The final answer is $\boxed{30^\circ}$.