= Solution
For a constant <electromagnetic field tensor>, the <relativistic Lorentz force> is a linear system with constant coefficients, so
$$
\boxed{u^\mu(\tau)=
\left[\exp\!\left(\frac qmF\tau\right)\right]^\mu{}_{\nu}u^\nu(0).}
$$
Put
$$
a=\frac{qE}{mc},
\qquad
b=\frac{qB}{m},
\qquad
\gamma_0=\frac1{\sqrt{1-v_0^2/c^2}}.
$$
The initial <four-velocity> is $(u^0,u^1,u^2,u^3)=\gamma_0(c,0,v_0,0)$. For parallel fields along $x$, the field tensor has one hyperbolic block in the $(0,1)$ plane and one rotation block in the $(2,3)$ plane. With the orientation convention in the question,
$$
u^0=\gamma_0c\cosh(a\tau),
\qquad
u^1=\gamma_0c\sinh(a\tau),
$$
$$
u^2=\gamma_0v_0\cos(b\tau),
\qquad
u^3=-\gamma_0v_0\sin(b\tau).
$$
Taking the initial spacetime point to be the origin and integrating $u^\mu=dx^\mu/d\tau$ gives the <relativistic motion in parallel electric and magnetic fields>
$$
\boxed{
\begin{aligned}
t(\tau)&=\frac{\gamma_0}{a}\sinh(a\tau),\\
x(\tau)&=\frac{\gamma_0c}{a}\bigl(\cosh(a\tau)-1\bigr),\\
y(\tau)&=\frac{\gamma_0v_0}{b}\sin(b\tau),\\
z(\tau)&=\frac{\gamma_0v_0}{b}\bigl(\cos(b\tau)-1\bigr).
\end{aligned}}
$$
The expressions have their continuous $a\to0$ or $b\to0$ limits when one field vanishes; reversing the sign of $q$ reverses the electric acceleration and magnetic rotation.
Eliminating <proper time> between the first two equations yields
$$
\boxed{x(t)=\frac{\gamma_0c}{a}
\left[\sqrt{1+\left(\frac{at}{\gamma_0}\right)^2}-1\right].}
$$
At early coordinate time,
$$
x(t)=\frac{qE}{2m\gamma_0}t^2+O(t^4),
$$
which is uniformly accelerated motion with the transverse relativistic inertia factor $\gamma_0$. At late positive time,
$$
x(t)=\operatorname{sgn}(qE)ct-\frac{\gamma_0mc^2}{qE}+O(t^{-1}),
$$
so the longitudinal speed approaches the <speed of light>.
The $y$-$z$ projection satisfies
$$
y^2+\left(z+\frac{\gamma_0v_0}{b}\right)^2
=\left(\frac{\gamma_0v_0}{b}\right)^2.
$$
It is therefore a circle of radius and proper-time period
$$
\boxed{R=\frac{\gamma_0m|v_0|}{|q|B},
\qquad
\Delta\tau=\frac{2\pi m}{|q|B}.}
$$
For the pitch magnitude assume $v_0>0$ and put $\alpha=|q|E/(mc)$, $\Omega=|q|B/m$, and
$$
r=\alpha\Delta\tau=\frac{2\pi E}{cB}.
$$
Number the $n$th proper-time period as $[(n-1)\Delta\tau,n\Delta\tau]$. Its longitudinal displacement is
$$
\Delta x_n=\frac{\gamma_0c}{\alpha}
\bigl[\cosh(nr)-\cosh((n-1)r)\bigr].
$$
Consequently, as $n\to\infty$,
$$
P_n=\frac{\Delta x_n}{R}
\sim\frac{c\Omega}{2\alpha v_0}(1-e^{-r})e^{nr}
=A\exp\!\left(\frac{2\pi En}{cB}\right),
$$
where
$$
\boxed{A=\frac{c^2B}{2Ev_0}
\left(1-e^{-2\pi E/(cB)}\right).}
$$
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