= Solution
Write $a_y=E^0_{A,y}$, $a_z=E^0_{A,z}$, $b_y=E^0_{B,y}$ and $b_z=E^0_{B,z}$, evaluated at the same transverse coordinates $(y,z)$. A superscript $0$ labels each filament's own <rest frame>, rather than a single common frame in all cases. The filaments are infinite and uniform along $x$, so the change of longitudinal coordinate under a <Lorentz boost> creates no extra dependence in these functions. After the acceleration transients have passed, the <magnetic field> of each filament vanishes in its own <rest frame>.
For a filament moving at $+v\widehat{\mathbf x}$, use the inverse of part (a): its transverse <electric field> in the laboratory is $\gamma\mathbf E^0$ and its <magnetic field> is $\mathbf v\times\mathbf E/c^2$. <Superposition> for the two co-moving filaments therefore gives, in $O$,
$$
\boxed{\begin{aligned}
E_x&=0,& B_x&=0,\\
E_y&=\gamma(a_y+b_y),&B_y&=-\frac{\gamma v}{c^2}(a_z+b_z),\\
E_z&=\gamma(a_z+b_z),&B_z&=\frac{\gamma v}{c^2}(a_y+b_y).
\end{aligned}}
$$
In the common co-moving frame $O'$ both filaments are at rest, so
$$
\boxed{\begin{aligned}
E'_x&=0,&B'_x&=0,\\
E'_y&=a_y+b_y,&B'_y&=0,\\
E'_z&=a_z+b_z,&B'_z&=0.
\end{aligned}}
$$
For example, $B'_y=\gamma[-\gamma v(a_z+b_z)/c^2+v\gamma(a_z+b_z)/c^2]=0$, a useful sign check. The supplied $\eta$ is the proper <linear charge density>. Thus each laboratory density is $\gamma\eta$ and each laboratory <electric current> is $\gamma\eta v$, consistently with <length contraction> and the <four-current> transformation. In $O'$ each density is $\eta$ and both currents vanish.
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