Let $ f : \mathbf{R} \to \mathbf{R} $ be a smooth function such that $f'(x)=f(1-x)$ for all $x$ and $f(0)=1$. Find $f(1)$.
**
   We substitute \( x = 1 \) into the general solution:
   \[
   f(1) = A \sin 1 + \cos 1
   \]
   Since \( f(1) = A \), we have:
   \[
   A = A \sin 1 + \cos 1
   \]
   Solving for \( A \):
   \[
   A (1 - \sin 1) = \cos 1
   \]
   \[
   A = \frac{\cos 1}{1 - \sin 1}
   \]

7. **Conclusion:**
   Therefore, the value of \( f(1) \) is:
   \[
   f(1) = \frac{\cos 1}{1 - \sin 1}
   \]

The final answer is \( \boxed{\frac{\cos 1}{1 - \sin 1}} \)