= Solution
Use the <Minkowski metric> $g^{\mu\nu}=\operatorname{diag}(1,-1,-1,-1)$ and <Feynman slash notation> $\not p=\gamma^\mu p_\mu$. The <Pauli matrices> satisfy $\sigma_j\sigma_k+\sigma_k\sigma_j=2\delta_{jk}I_2$. In the <Dirac representation of the gamma matrices>, direct block multiplication gives
$$
(\gamma^0)^2=I_4,\qquad \gamma^0\gamma^j=\begin{pmatrix}0&\sigma_j\\\sigma_j&0\end{pmatrix}=-\gamma^j\gamma^0,
$$
and
$$
\gamma^j\gamma^k=\begin{pmatrix}-\sigma_j\sigma_k&0\\0&-\sigma_j\sigma_k\end{pmatrix}.
$$
Thus the spatial <anticommutators> are $-2\delta_{jk}I_4$, proving \b[the Clifford relations]
$$
\boxed{\{\gamma^\mu,\gamma^\nu\}=2g^{\mu\nu}I_4.}
$$
Since partial derivatives commute, their symmetric product selects the symmetric part of the <gamma matrices>. Multiplying the <Dirac equation> by $i\gamma^\mu\partial_\mu+m$ gives
$$
(i\not\partial+m)(i\not\partial-m)\psi
=(-\gamma^\mu\gamma^\nu\partial_\mu\partial_\nu-m^2)\psi
=-(\Box+m^2)\psi=0.
$$
Hence \b[every spinor component satisfies the Klein-Gordon equation] with the same mass.
For <parity>, let $x_P=(t,-\mathbf x)$ and $\psi_P(x)=\gamma^0\psi(x_P)$. The chain rule gives $\partial_0\psi(x_P)=(\partial_0\psi)(x_P)$ and $\partial_j\psi(x_P)=-(\partial_j\psi)(x_P)$. Combining these signs with $\gamma^j\gamma^0=-\gamma^0\gamma^j$ proves
$$
(i\not\partial-m)\psi_P(x)
=\gamma^0[(i\not\partial-m)\psi](x_P)=0.
$$
This establishes <parity symmetry in quantum field theory> for the free <Dirac equation>, rather than merely asserting its covariance.
A positive-energy <plane wave> $\psi=u(p)e^{-ip\cdot x}$ solves $(\not p-m)u=0$. Write $u=(\chi,\eta)^T$. The two block equations are
$$
(E_p-m)\chi-(\boldsymbol\sigma\cdot\mathbf p)\eta=0,\qquad
(\boldsymbol\sigma\cdot\mathbf p)\chi-(E_p+m)\eta=0.
$$
The second gives $\eta=(\boldsymbol\sigma\cdot\mathbf p)\chi/(E_p+m)$. In the first, $(\boldsymbol\sigma\cdot\mathbf p)^2=\mathbf p^2 I_2$ and $E_p^2-\mathbf p^2=m^2$ make the remaining coefficient zero. There are two independent two-component choices $\chi_s$, so
$$
\boxed{u_s(p)=\begin{pmatrix}\chi_s\\(\boldsymbol\sigma\cdot\mathbf p)\chi_s/(E_p+m)\end{pmatrix},\qquad \psi_{p,s}(x)=u_s(p)e^{-ip\cdot x}.}
$$
For a massive <Dirac spinor>, choose $\chi_s$ as the two rest-frame <spin> states along a fixed axis, for example $\sigma_3\chi_s=\pm\chi_s$. The label $s$ distinguishes these canonical <spin> states after boosting; it is not automatically a <helicity> label.
Now $p\cdot x_P=p_P\cdot x$ and $\gamma^0u_s(p)=u_s(p_P)$ for the same fixed $\chi_s$. Therefore
$$
\boxed{\psi_{p,s}(x)\longmapsto\psi_{p_P,s}(x).}
$$
The <parity action on canonical massive spin states> leaves the canonical <spin> label unchanged while reversing the <momentum>. <Spin> is an axial vector under <parity>; its orientation is unchanged. Consequently <helicity>, the projection of <spin> along <momentum>, reverses. If $s$ had instead been chosen to denote <helicity>, the transformed label would have the opposite sign.
The <Dirac adjoint> is $\bar\psi=\psi^\dagger\gamma^0$, with $\gamma^{0\dagger}=\gamma^0$. Thus
$$
\bar\psi_P(x)=[\gamma^0\psi(x_P)]^\dagger\gamma^0=\psi^\dagger(x_P)=\bar\psi(x_P)\gamma^0.
$$
The <chirality matrix> has $\gamma^5=i\gamma^0\gamma^1\gamma^2\gamma^3=\begin{pmatrix}0&I_2\\I_2&0\end{pmatrix}$ in this representation, so $(\gamma^5)^2=I_4$ and it anticommutes with every $\gamma^\mu$. In particular $\gamma^0\gamma^5\gamma^0=-\gamma^5$. The <parity transformation of a Dirac bilinear> therefore yields
$$
\boxed{\bar\psi_P(x)\psi_P(x)=\bar\psi(x_P)\psi(x_P),\qquad
\bar\psi_P(x)\gamma^5\psi_P(x)=-\bar\psi(x_P)\gamma^5\psi(x_P).}
$$
The first bilinear is a scalar under <parity>, whereas the second is a <pseudoscalar>.
For the requested <gamma matrix trace identities>, use the <cyclic property of the trace>. Conjugation of a product of $n$ <gamma matrices> by $\gamma^5$ multiplies it by $(-1)^n$, but leaves its <trace> unchanged. All odd products therefore have zero <trace>. Taking the <trace> of the <Clifford algebra> relation and using $\operatorname{Tr}I_4=4$ gives $\operatorname{Tr}(\gamma^\alpha\gamma^\beta)=4g^{\alpha\beta}$. For four factors, move the first <gamma matrix> successively past the next three using the <anticommutators>. If $A=\operatorname{Tr}(\gamma^\alpha\gamma^\beta\gamma^\rho\gamma^\delta)$, this gives
$$
A=2g^{\alpha\beta}\operatorname{Tr}(\gamma^\rho\gamma^\delta)
-2g^{\alpha\rho}\operatorname{Tr}(\gamma^\beta\gamma^\delta)
+2g^{\alpha\delta}\operatorname{Tr}(\gamma^\beta\gamma^\rho)-A.
$$
Thus $A=4(g^{\alpha\beta}g^{\rho\delta}-g^{\alpha\rho}g^{\beta\delta}+g^{\alpha\delta}g^{\beta\rho})$. Contracting with the relevant <four-vectors> proves all four answers:
$$
\boxed{\begin{aligned}
\operatorname{Tr}(\not p)&=0,\\
\operatorname{Tr}(\not p\not q)&=4p\cdot q,\\
\operatorname{Tr}(\not p\not q\not k)&=0,\\
\operatorname{Tr}(\not p\gamma^\mu\not q\gamma^\nu)&=4[p^\mu q^\nu+p^\nu q^\mu-(p\cdot q)g^{\mu\nu}].
\end{aligned}}
$$
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