The reciprocal vector [hkl] is perpendicular to the planes described by the Miller indices (hkl). This is a general relationship, not specific to cubic systems.
So yes, it works for hexagonal too.
Same is true for the relationship that gives the spacing of the planes as the inverse of the magnitude of the corresponding reciprocal vector (up to a factor of $ 2 \pi $). If you look up the proofs of these relationships you will see that no assumption about the crystal symmetry is made.
This is more a wording problem than a physics problem.
In order to solve this, consider a simple harmonic oscillator, i.e. a particle of mass $m$ on a weightless spring with spring constant $D$. This system has an eigenfrequency
$f_0 = \frac{1}{2\pi}\sqrt{\frac{D}{m}}$, which is also called resonance frequency.
If we excite a harmonic oscillator, it always oscillates with the frequency of the excitation. Thus, if we apply an external force $F(t) = F_0 \, \cos(\omega t)$ the harmonic oscillator oscillates with the frequency $f_{\textrm{force}} = \frac{\omega_{\textrm{force}}}{2\pi} = \frac{\omega}{2\pi}$. The amplitude of the oscillation is substantially influence by the difference $\Delta f = f_{\textrm{force}} - f_0$. If this difference is zero, the amplitude of the oscillation is maximal. So, if somebody says "if the excitation frequency is in resonance with the oscillator", it means that the excitation frequency is equal to the resonance frequency of the oscillator. Therefore, the amplitude of the oscillator is maximal.
Now, let's consider your problem. You have two waves which propagate through space. At each point in space these two waves add together. Thus, the total amplitude is given by
\begin{align}
A_{tot}(t, x) &= A_1(t, x) + A_2(t, x)
\\
&= A_1 \cdot \cos(k x - \omega t+ \varphi_{1}(x)) + A_2 \cdot \cos(k x - \omega t + \varphi_{2}(x))
\end{align}
I write it in 1D, but this is true generally -- I just don't won't to write all the vectors. If we consider a certain point in space and in time, the expression $k x - \omega t$ is the same for both waves. Thus, this oscillating term is not of interest and we drop it -- mathematically, you can use trigonometric identities to get a common pre-factor. Thus, if we split the amplitude and only consider the common amplitude $A = \min\{A_1, A_2\}$ of each term, we obtain
$$
A_{common} = A \cdot \cos(\varphi_{1}(x))
+ A \cdot \cos(\varphi_{2}(x))
$$
Now, without loss of generality, we can choose $\varphi_1(x) = 0$ and $\varphi_2(x) = \Delta \varphi(x) = \varphi_{2}(x) - \varphi_{1}(x)$. This simplifies the relation to
$$
A_{common} = A \big[
1
+ \cos(\Delta \varphi)
\big]
$$
Hence, the phase difference $\Delta \varphi$
determines the maximal amplitude of the oscillation. Some people say, the two waves are in resonance if their relative phase difference vanishes and thus the amplitude of oscillation is maximal.
The last missing piece for your answer is, that we often imagine that space is made up from invisible oscillators.
The two waves propagate through space independently, except at the points of the lattice, where they resonate with the oscillators, so their phase must coincide.
It means, that two independent waves (as described above) satisfy $\Delta \varphi = n \cdot 2\pi$ at the positions in space, where they interfere constructively. However, instead of using this formulation, the author imagines invisible oscillators in space. Now, at the positions of constructive interferences, the invisible oscillator gets excited by the two independent waves. When these two waves are in phase (which means that their relative phase difference is $\Delta \varphi = n \cdot 2\pi$) they interfer constructively. Thus, the resulting amplitude of the harmonic oscillator is at it maximum.
Best Answer
The distinction between the two indices is quite subtle.Perhaps it is better to start from the laue condition. Let's say I have rows of atoms with spacing $a$. We are in the fraunhofer limit so when two x-ray beams come in and scatter off neighboring atoms they come in parallel and go out parallel. We can then consider the path length difference between the incoming beams and out going beams i.e $ \Delta^1 = \Delta_1 - \Delta_2= a \cos \alpha_1 - a \cos \alpha_2 $ where $\alpha_1$ is the angle between the incoming beam and the row of atoms and $\alpha_2$ is the angle between the outgoing beam and the row of atoms.The condition for constructive interference is that $\Delta^1 = h \lambda$ where $\lambda$ is the wavelength and h is an integer. This analysis can be done in all three directions and then the condition has to be true in all three directions giving $\Delta^2 = k\lambda \text{ and } \Delta^3 = l\lambda $.(There will be corresponding angles for the other path length differences).
So we can define incoming unit vectors $s_1 = (\cos \alpha_1,\cos \beta_1, \cos \gamma_1)$ and outgoing $s_2 = (\cos \alpha_2, \cos \beta_2, \cos \gamma_2)$. From this we consider the scatter wave $s_1-s_2= G \lambda $ with $G = \frac{1}{a}(h,k,l)$ since $s_1 ,s_2$ are unit vectors $G$ will be perpendicular to the vector that bisects the angle between $s_1 $ and $s_2$.
Now comes the first crucial point, draw planes that are parallel to the plane that bisects the angle between the incoming and outgoing beams. These are the lattice planes and they define the miller indices.Laue indices will be defined by another set of planes. Back to the lattice planes , call the distance between each successive plane $d_{hkl}$ this is the distance that appears in Bragg's law.When we pick a primitive cell then (hkl)are our miller indices.
What about laue indices? Well we look at the bragg's law $n \lambda = 2 d_{hkl} \sin \theta_n $ . Now comes the second crucial point, note that the $m^{th}$ order reflection off the (hkl) plane can be regarded as $n^{th}$ order if I $ \textit{define}$ new planes at distance $\frac{d_{hkl}n}{m}$ i.e $ m \frac{n}{m}\lambda=2 (d_{hkl}\frac{n}{m})\sin \theta_m $ This family of planes is then given by the notation $ \frac{m h}{n} \frac{m k}{n} \frac{m l}{n}$ without parantheses. These are the laue indices, note how they can have common factors even if we started off with a primitive cell.