= Solution
For this question use the <mostly-plus Dirac convention>: the <Minkowski metric> is $\eta_{ab}=\operatorname{diag}(-1,1,1,1)$ and
$$
\boxed{\{\gamma^a,\gamma^b\}=-2\eta^{ab}I_4},\qquad
\boxed{(i\gamma^a\partial_a+m)\Psi=0}.
$$
This convention matches the printed plane-wave phase and final identity. It is related to the usual mostly-minus <Dirac equation> by reversing the metric and taking the negatives of the usual <gamma matrices>. In particular, the resulting <Dirac action> and its <Dirac adjoint> describe the same physical massive field. The <Dirac gamma matrices> are four complex $4\times4$ matrices representing the spacetime <Clifford algebra> with quadratic form $-\eta$. The irreducible complex representation has dimension four. A convenient explicit choice is the negative of the standard <Dirac representation of the gamma matrices>:
$$
\gamma^0=-\begin{pmatrix}I_2&0\\0&-I_2\end{pmatrix},\qquad
\gamma^i=-\begin{pmatrix}0&\sigma^i\\-\sigma^i&0\end{pmatrix},\qquad i=1,2,3,
$$
where $\sigma^i$ are the <Pauli matrices>. Their multiplication law verifies the displayed <anticommutator>.
In this representation the <gamma matrix adjoint and transpose identities> are
$$
(\gamma^0)^\dagger=\gamma^0,\qquad(\gamma^i)^\dagger=-\gamma^i,\qquad
\boxed{(\gamma^a)^\dagger=\gamma^0\gamma^a\gamma^0},
$$
and
$$
(\gamma^0)^T=\gamma^0,\quad(\gamma^1)^T=-\gamma^1,\quad
(\gamma^2)^T=\gamma^2,\quad(\gamma^3)^T=-\gamma^3.
$$
The invariant way to express the latter pattern uses the <charge-conjugation matrix>:
$$
C=i\gamma^2\gamma^0,\qquad C^T=-C,\qquad C^\dagger=C^{-1},\qquad
\boxed{C^{-1}\gamma^aC=-(\gamma^a)^T}.
$$
Individual transpose signs depend on the basis. More generally, a <similarity transformation> $\gamma'^a=M\gamma^aM^{-1}$ changes the <Hermitizing matrix> to $H'=M^{-\dagger}\gamma^0M^{-1}$ and the <charge-conjugation matrix> to $C'=MCM^T$. Then $\gamma'^{a\dagger}=H'\gamma'^aH'^{-1}$ and $C'^{-1}\gamma'^aC'=-\gamma'^{aT}$. Thus the simple formula with $\gamma'^0$ itself presumes a compatible Hermitian basis, rather than an arbitrary nonunitary <similarity transformation>.
Applying $(i\gamma^a\partial_a-m)$ to the <Dirac equation> gives $(\Box-m^2)\Psi=0$. Its <mass shell> is $p^2=-m^2$, so the frequencies are $p^0=\pm\sqrt{\mathbf p^2+m^2}$. The <Dirac spinor> transforms in the four-component <Spinor representation of the Lorentz group>. Under spatial rotations, the two upper and the two lower components each transform as a two-component spin-$1/2$ representation: the <spin angular momentum> matrices are $\operatorname{diag}(\sigma^i,\sigma^i)/2$. At rest the positive-energy equation selects the upper two components, giving two independent <spin> polarizations, and the negative-frequency equation selects the lower two.
In the quantum theory a mode expansion is
$$
\Psi(x)=\sum_{s=1}^2\int\frac{d^3p}{(2\pi)^3\sqrt{2E_{\mathbf p}}}
\left[b_s(\mathbf p)u_s(p)e^{ip\cdot x}+d_s^\dagger(\mathbf p)v_s(p)e^{-ip\cdot x}\right],\qquad p^0=E_{\mathbf p}>0.
$$
With the <mostly-plus Dirac convention>, $e^{ip\cdot x}=e^{-iEt+i\mathbf p\cdot\mathbf x}$ is positive frequency. The negative-frequency coefficient obeys $(\not p+m)v_s(p)=0$. The <fermionic annihilation operators> $b_s$ and $d_s$ satisfy the <canonical anticommutation relations>, with their respective <fermionic creation operators>. The $b_s^\dagger$ excitations are particles; the $d_s^\dagger$ excitations are <antiparticles> with the same positive <energy>, <mass> and spin-$1/2$ but opposite charge. After <normal ordering>, the <Hamiltonian operator> contains positive multiples of $b_s^\dagger b_s+d_s^\dagger d_s$. Reinterpreting the negative-frequency part as antiparticle creation supplies a spectrum bounded below rather than a physical tower of negative-energy particles.
For the printed $e^{ip\cdot x}$ wave, $\partial_a\Psi=ip_a\Psi$. Substitution gives
$$
(-\gamma^ap_a+m)u_s(p)=0,\qquad
\boxed{(\not p-m)u_s(p)=0},\qquad\not p=\gamma^ap_a.
$$
The spin label $s$ indexes the two states of a spin-$1/2$ particle, rather than varying the particle's total spin. For real on-shell $p$, <Hermitian conjugation> and $\gamma^{a\dagger}=\gamma^0\gamma^a\gamma^0$ give
$$
u_s^\dagger(p)(\not p^\dagger-m)=0
\quad\Longrightarrow\quad
\boxed{\bar u_s(p)(\not p-m)=0},\qquad\bar u_s=u_s^\dagger\gamma^0.
$$
These are right and left null-vector equations for the same on-shell matrix.
To obtain the <Gordon identity>, take both external <Dirac spinors> to have the same real <mass> $m$. Their two equations imply
$$
\bar u_{s'}(p')\bigl[\gamma^a\not p+\not p'\gamma^a-2m\gamma^a\bigr]u_s(p)=0.
$$
Define $\gamma^{ab}=[\gamma^a,\gamma^b]/2$. The <Clifford algebra> relation yields
$$
\gamma^a\gamma^b=-\eta^{ab}+\gamma^{ab},\qquad
\gamma^b\gamma^a=-\eta^{ab}-\gamma^{ab}.
$$
Therefore
$$
\gamma^a\not p+\not p'\gamma^a
=-(p^a+p'^a)-\gamma^{ab}(p'_b-p_b).
$$
Multiplying the previous null-vector equation by $-1$ proves the required formula exactly:
$$
\boxed{\bar u_{s'}(p')\gamma^{ab}(p'_b-p_b)u_s(p)
+\bar u_{s'}(p')(2m\gamma^a+p^a+p'^a)u_s(p)=0}.
$$
For $m\ne0$ it can be solved for the vector-current matrix element, separating a momentum term from the antisymmetric <Dirac spinor> term. The identity before division also holds at $m=0$. \b[The signs depend jointly on the metric, Clifford relation, Dirac mass term and plane-wave phase.] In the mostly-minus convention of Questions 1 and 3, the printed phase instead gives $(\not p+m)u=0$; the corresponding identity uses $(p_b-p'_b)$ in the antisymmetric term. Mixing that convention with the formula proved here would produce an apparent sign error.
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