= Solution
The vacuum <Maxwell equations> are $\nabla\cdot E=\nabla\cdot B=0$, $\nabla\times E=-B_t$ and $\nabla\times B=E_t/c^2$. Taking the curl of Faraday's equation and using $\nabla\times(\nabla\times E)=\nabla(\nabla\cdot E)-\nabla^2E$ gives $-\nabla^2E=-E_{tt}/c^2$. The same calculation for Ampère's equation gives $\nabla^2B=B_{tt}/c^2$. Hence \b[each Cartesian component satisfies the <electromagnetic wave equation>].
Model the walls as a <perfect conductor>, with no oscillating fields in its interior. The tangential <electric field> is zero at the surface, otherwise it would drive unbounded conductor currents. The normal <magnetic field> is zero, since the divergence-free field has a continuous normal component and the interior oscillating field is zero. Thus $n\times E=0$ and $n\cdot B=0$. Normal $E$ and tangential $B$ may be nonzero because surface charge and surface current support their jumps.
The PDF writes the full <magnetic field> as $B=B_0(x,y)\hat z e^{i(kz-\omega t)}$. Taken literally, this cannot be a nonzero travelling wave: $\nabla\cdot B=ikB_z$, so for $k\ne0$ it forces $B_z=0$. The subsequent requests require transverse components. The consistent intended reading is \b[$B_z=B_0(x,y)e^{i(kz-\omega t)}$], with transverse fields determined by the <Maxwell equations>. Use that reading below.
All fields have the same $z,t$ factor. Put $\Delta=(\omega/c)^2-k^2$. The harmonic <Maxwell equations> are $\nabla\times E=i\omega B$ and $\nabla\times B=-i\omega E/c^2$. Their $x$ and $y$ components give
$$
E_y=-\frac{c^2k}{\omega}B_x-\frac{ic^2}{\omega}\partial_xB_z,\qquad
E_x=\frac{c^2k}{\omega}B_y+\frac{ic^2}{\omega}\partial_yB_z,
$$
together with $\partial_yE_z-ikE_y=i\omega B_x$ and $ikE_x-\partial_xE_z=i\omega B_y$. Substitution and solving, for $\Delta\ne0$, gives
$$
\boxed{B_x=\frac i\Delta\left(k\partial_xB_z-\frac\omega{c^2}\partial_yE_z\right),\qquad
B_y=\frac i\Delta\left(k\partial_yB_z+\frac\omega{c^2}\partial_xE_z\right).}
$$
The corresponding electric components are
$$
E_x=\frac i\Delta(k\partial_xE_z+\omega\partial_yB_z),\qquad
E_y=\frac i\Delta(k\partial_yE_z-\omega\partial_xB_z).
$$
For a <transverse electric mode of a rectangular waveguide>, $E_z=0$. The <electromagnetic wave equation> becomes
$$
(\partial_x^2+\partial_y^2+\Delta)B_0=0.
$$
At $x=0,a$, the tangential component $E_y=-i\omega\partial_xB_z/\Delta$ vanishes, so $\partial_xB_0=0$; at $y=0,b$, the corresponding condition $E_x=0$ gives $\partial_yB_0=0$. These are <Neumann boundary conditions>, and the normal-$B$ conditions give the same conclusions through the displayed transverse magnetic components. Separation gives
$$
B_0=C_{mn}\cos\frac{m\pi x}{a}\cos\frac{n\pi y}{b},\qquad
\Delta=\kappa_{mn}^2=\left(\frac{m\pi}{a}\right)^2+\left(\frac{n\pi}{b}\right)^2,
$$
where $m,n\ge0$ are integers and $(m,n)\ne(0,0)$. The constant candidate is not a physical oscillating mode: its transverse derivatives vanish, so it has $E=0$ and Faraday's equation forces $B=0$ for $\omega\ne0$. More generally a hollow single-conductor guide has no transverse-electromagnetic zero-cutoff mode: its transverse electrostatic potential is harmonic with the same constant boundary value, hence constant, leaving zero transverse field.
The dispersion relation is $\omega^2=c^2(k^2+\kappa_{mn}^2)$. The <waveguide cutoff frequency> for this mode is $\omega_{mn}=c\kappa_{mn}$; below it $k$ is imaginary and the field is evanescent. Since $a\ge b$, the lowest cutoff occurs at $(m,n)=(1,0)$:
$$
\boxed{\omega_c=\frac{\pi c}{a},\qquad f_c=\frac c{2a}.}
$$
For a mode propagating in the positive $z$ direction ($k>0$), above its own cutoff $\omega_{mn}$, the <phase velocity> and <group velocity> are
$$
\boxed{v_{\mathrm{ph}}=\frac\omega k=\frac c{\sqrt{1-(\omega_{mn}/\omega)^2}}>c,\qquad
v_g=\frac{d\omega}{dk}=c\sqrt{1-(\omega_{mn}/\omega)^2}<c.}
$$
These <waveguide phase and group velocities> have product $c^2$. The superluminal phase speed describes motion of a constant phase, while a narrow wave packet travels at the subluminal <group velocity>; it does not transmit information faster than light. At cutoff $k=0$ there is no travelling mode with nonzero longitudinal flux.
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