 Solution

The automorphism $a$ cyclically permutes the first letter and leaves the remaining suffix unchanged, so $a^3=1$ and $a\ne 1$.

In the section notation,Thus $b^3=(a^3,1,b^3)=(1,1,b^3)$. An automorphism $c$ satisfying $c=(1,1,c)$ fixes every finite word: repeatedly entering the third subtree eventually reaches the end of the word. Hence $b^3=1$. Since $b(00)=01$, it is nonidentity, and both $a$ and $b$ have order three.