Let $a,b\in D_4$ and such that $ba=a^{3}b$. Prove that if $0\leq i<4$ and $0\leq j<2$, then $a^ib^ja^{i_1}b^{j_1}=a^{i+3i_1}b^{j+j_1}$,(where $i,j,i_1,j_1\in \mathbb{Z}$) also it is given $o(a)=4$ and $o(b)=2$. $D_4$ is a dihedral group of order $8$.
I could only verify the fact that if $ba=a^3b$, then $\forall i\in \mathbb {Z}$, $ba^i=a^{3i}b$. This is because $ba^i=baa^{i-1}=a^3ba^{i-1}=a^3baa^{i-2}=a^6ba^{i-2}$. Then, continuing on this way, we have, $a^{3i}b=ba^i$. But I dont know how to prove $a^ib^ja^{i_1}b^{j_1}=a^{i+3i_1}b^{j+j_1}$?
If in a group $G$ for some elements $a,b$ the relation $ba=a^3b$ holds, then the equality $$ b^ja^k=a^{3^jk}b^j\tag1 $$ is valid. To prove this equality, you can reason like this: $$ b^ja^k=b^{j-1}a^{3k}b=b^{j-2}a^{3^2k}b^2=\ldots. $$ One can also use the fact that the mapping $x\to bxb^{-1}$ is an automorphism of the group $G$ to prove $(1)$.
If we additionally know that $a^4=1$, then equality $(1)$ turns into $$ b^ja^k=a^{(-1)^jk}b^j.\tag2 $$ Finally, if we also know that $j=0$ or $j=1$, then we don't need either formula $(1)$ or formula $(2)$. We only need the formula $$ ba^k=a^{-k}b. $$
No one but you can tell exactly what your problem is. And only you can formulate it.