= Solution
With $\eta=\operatorname{diag}(1,-1)$, the <Clifford algebra> gives $(\gamma^0)^2=1$, $(\gamma^1)^2=-1$, and $\gamma^0\gamma^1=-\gamma^1\gamma^0$. Hence
$$
\gamma^5\gamma^0=-\gamma^1,\quad \gamma^0\gamma^5=\gamma^1,\qquad
\gamma^5\gamma^1=-\gamma^0,\quad \gamma^1\gamma^5=\gamma^0,
$$
so $\{\gamma^5,\gamma^0\}=\{\gamma^5,\gamma^1\}=0$. Its square is
$$
(\gamma^5)^2=\gamma^0\gamma^1\gamma^0\gamma^1
=-(\gamma^0)^2(\gamma^1)^2=1.
$$
Choose the following <gamma matrices>, expressed using the <Pauli matrices>:
$$
\boxed{\gamma^0=\sigma_1=\begin{pmatrix}0&1\\1&0\end{pmatrix},\qquad
\gamma^1=-i\sigma_2=\begin{pmatrix}0&-1\\1&0\end{pmatrix},\qquad
\gamma^5=\sigma_3=\begin{pmatrix}1&0\\0&-1\end{pmatrix}.}
$$
These satisfy $(\gamma^0)^\dagger=\gamma^0$, $(\gamma^1)^\dagger=-\gamma^1$ and $(\gamma^5)^\dagger=\gamma^5$. The <chirality> <eigenvalues> are therefore $\boxed{+1,-1}$, and we order the <spinor> as $\psi=(\psi_R,\psi_L)^{\mathsf T}$ throughout. Its <Dirac adjoint> is $\bar\psi=(\psi_L^\dagger,\psi_R^\dagger)$. No additional factor of $i$ multiplies $\gamma^0\gamma^1$ in this spacetime dimension. This realizes a <Dirac field in two spacetime dimensions>.
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