Based on my answer to this: Fermionic occupation operator and nearest neighbor Fermionic hopping interaction as a qubit operator, you can see that we have:
\begin{align}
\hat{a}_i &= \frac{1}{2} Z^{\otimes (i-1)} (X - iY),\\
\hat{a}_i^\dagger &=\frac{1}{2} Z^{\otimes (i-1)} (X + iY).\\
\end{align}
If $i=j$ we have:
\begin{align}
\{\hat{a}_i,\hat{a}_i^\dagger\} &\propto \frac{1}{4} ((X - iY)(X + iY) + (X + iY)(X - iY)),\\ &= \frac{1}{4} (X^2 + iXY + - iXY + Y^2 + X^2 -iXY + iXY + Y^2) \\
&=\frac{1}{4}(2X^2 + 2Y^2)\\
& =\frac{1}{4}(4I) \\
& = I.
\end{align}
All $Z$ operators are replaced by $I$ operators since $Z \times Z = I$, and this is also what was done in the penultimate step, where $X^2 = I$ and $Y^2 = I$ were used.
For the other anti-commutators we have:
\begin{align}
\{\hat{a}_i,\hat{a}_j\} &= \frac{1}{4} ((X - iY)(X - iY) + (X - iY)(X - iY)),\\ \{\hat{a}^\dagger_i,\hat{a}^\dagger_j\} &= \frac{1}{4} ((X + iY)(X + iY) + (X + iY)(X + iY)).
\end{align}
You can then do the same type of arithmetic that I went through in detail for the first anti-commutator.