If $a,b \in$ group $G$ such that $a^2=e, a*b^4*a=b^7$, prove that $b^{33}=e$

For any integer $n$ we have $$(ab^4a)^n=b^{7n}$$ or $$ab^{4n}a=b^{7n}$$ For $n=4$, $$ab^{16}a=b^{28}$$ and for $n=7$, $$ab^{28}a=b^{49}$$ Therefore $$b^{16}=b^{49}$$

EDIT: With same technique, it can be shown that if $a^2=e$ and $ab^r=b^sa$ then $b^{r^2}=b^{s^2}$.

Note that \begin{align} (aba^{-1})^4&=b^7\text{, and }\\ b^4&=(aba^{-1})^7\nonumber\\ (aba^{-1})^7&=b^4. \text{So 'dividing' the first equality by the last,}\\ (aba^{-1})^{-3}&=b^{3}. \text{ Multiplying it with the first equality,}\\ aba^{-1}&=b^{10}\\ (aba^{-1})^4&=b^{40}=b^7\\ \text{So }b^{33}&=e \end{align}