= Solution
Distinct <gamma matrices> anticommute, and $(\gamma^0)^2=I$, $(\gamma^j)^2=-I$ for $j=1,2,3$. Moving a particular $\gamma^\mu$ through the other three factors in $\gamma^5$ gives three minus signs. Therefore
$$
\boxed{\{\gamma^5,\gamma^\mu\}=0.}
$$
Pairing equal factors in the square involves $3+2+1=6$ interchanges. The <Clifford algebra> consequently gives
$$
\boxed{(\gamma^5)^2=(-1)^6\prod_{\mu=0}^3(\gamma^\mu)^2=-I.}
$$
In particular $(\gamma^5)^{-1}=-\gamma^5$; replacing it by the conventional <chirality> matrix would change this answer.
The <trace> is unchanged by a <similarity transformation>. Since $(\gamma^0)^{-1}=\gamma^0$ and $\gamma^0\gamma^5\gamma^0=-\gamma^5$, <cyclic property of the trace> gives $\operatorname{tr}\gamma^5=-\operatorname{tr}\gamma^5$. Also
$$
\operatorname{tr}(\gamma^5\gamma^\mu)=\operatorname{tr}(\gamma^\mu\gamma^5)=-\operatorname{tr}(\gamma^5\gamma^\mu).
$$
Thus
$$
\boxed{\operatorname{tr}\gamma^5=\operatorname{tr}(\gamma^5\gamma^\mu)=0.}
$$
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