Assuming $ab^2 = b^3a$ and $a^2=1$ prove that the order of $b$ is $5$.

Note that $$b^5 = b^3a^2b^2=(b^3a)(ab^2)=(b^3a)(b^3a)=b^3(ab^2)ba=b^3(b^3a)ba=b^6aba,$$ multiplying by $b^{-5}$ from the left we get $1=baba$. Multiplying by $a$ from the right we get $a=baba^2=bab$. Thus $$1=a^2=(bab)^2=b(ab^2)ab=b(b^3a)ab=b^4a^2b=b^4\cdot 1 \cdot b = b^5.$$

Consider that $$ab^2a=b^3$$ And so: $$b^2=ab^3a=ab^2ba=ab^2aaba=b^3aba$$ From here: $$b^{-1}=aba$$ Thus $$b^{-2}=ab^2a=b^3$$ which means: $$b^5=1$$