Why are coherent states of harmonic oscillators called "coherent"? Why are coherent states of the harmonic oscillator called coherent? Coherent in what sense? Why are these states so special/useful? 
From Wikipedia: 

In physics, two wave sources are perfectly coherent if they have a constant phase difference and the same frequency. 

 A: Coherent States of Harmonic Oscillator are Quantum Mechanical states which have definite phase and Minimum Uncertainty.  The Quantum Mechanical states of Harmonic Oscillator do not have definite phase (i.e,one photon being in phase with another photon is not valid in Quantum mechanics) so we need a Quantum Mechanical state to model a coherent laser light (which is thought of many light sources in phase with each other).
The Coherent state of Harmonic Oscillator can be obtained by a displacement operator acting on vaccum state
$$ D|0\rangle=|\alpha\rangle$$ 
Where $$D=e^{(\alpha{a}^{\dagger}-\alpha^*a)}$$
The Coherent State is defined as,
$$|\alpha\rangle=e^{\frac{-|\alpha|^2}{2}\sum_{n=0}^{\infty}}\frac{\alpha^n}{\sqrt{n!}}|n\rangle$$
Where $\alpha=|\alpha|e^{i\theta}$ is a complex number with '$\theta$' the phase of the coherent state.
The Coherent State obey Uncertainty relation,
$$\Delta{x}\Delta{y}=\frac{\hbar}{2}$$
The Ground state of HO do not have any definite phase, so the gaussian is centered at origin in phase space.  But the Coherent state has a definite phase so the gaussian is not necessarily centered at the origin. 


In Quantum Optics context '$x$' and '$y$' are related with Electric and Magnetic fields.  The Idea of Coherent state is extensively used in Quantum Optics. 
A: In my opinion, the most intuitive way of explaining the meaning of harmonic oscillator coherent state is the following:
An Harmonic Oscillator Coherent State (AOCS) is a solution of the time-dependent Schodinger equation for the quantum harmonic oscillator. This means that the probability density distribution, $|\psi|^2(x,t)$ evolves in time, i.e. it is not a stationary state. Do you remember the ground state of an harmonic oscillator?! The famous $|0\rangle$ state corresponds to a Gaussian probability density, centered in the origin. The particular feature of this state is that the uncertainity on the position, $\Delta x$ and the uncertainity on the momentum $\Delta p$ minimize Heisemberg's uncertainity principle. Eigenstate $|0\rangle$ is a stationary state and the associated probability density $|\psi|^2(x)$ doesn't depend on time. 
Well, an Harmonic Oscillator Coherent state can be visualized as an "oscillating" $|0\rangle$ state. In other words, an AOCS is a gaussian wavefunction which oscillates back and forth, around position $x=0$. But at every instant the shape of the gaussian is always the same (and minimizes the uncertainity principle). So, some quantities like $\langle x \rangle $ (which corresponds to the peak of the Gaussian wavefunction), $\langle p \rangle $, (which can be associated to the velocity of the gaussian's peak) have a sinusoidal time evolution. That's why they often go under the name of semi-classical quantum states! In other words: if the amplitude of oscillations is $\gg$  $\Delta x$, you re-obtain the classical equations of motion for a point-like massive particle subjected to an elastic force. 
Of course you have many possible coherent states. To be more precise, coherent states form a 2D planar manifold. This means that, in order to build an AOCS you have 2 degree of freedom: amplitude and phase of the oscillation of the gaussian wave packet. 
They are coherent in the sense that the phase of the various chromatic components is arranged in such a way to reproduce this oscillating behaviour.
A: Coherent states are eigenvectors for the (bosonic) annihilator,$$\hat a ~|\alpha\rangle = \alpha~|\alpha\rangle,$$and if we define the position and momentum quadratures as $\hat x = \hat a^\dagger + \hat a,$ $\hat p = i \hat a^\dagger - i \hat a,$ we have $[\hat x, \hat p] = 2i$ and the dimensionless Hamiltonian $\hbar\omega ~ \hat a^\dagger \hat a = \frac12 \hbar\omega~x^2 + \frac12 \hbar\omega~p^2 + \text{const.}$ to guide us. We can immediately see that in the coherent state we have $\langle x \rangle = \alpha^* + \alpha = 2 ~\Re~{\alpha}$ whereas $\langle p \rangle = i~\alpha^* - i ~ \alpha = 2 ~\Im~\alpha, $ so the position and momentum plane is basically just the complex plane $\mathbb C$ that $\alpha$ lives on.
Now this Hamiltonian of course has an eigenbasis $\hat a^\dagger \hat a ~ |n\rangle = n ~|n\rangle$ and in terms of that basis we see a recurrence that if $|\alpha\rangle = \sum_n c_n |n\rangle$ then we can work out that $\alpha |\alpha\rangle = \hat a |\alpha\rangle$ implies $$c_n \sqrt{n} = \alpha c_{n-1},\text { so } c_n = c_0 \frac{\alpha^n}{\sqrt{n!}}.$$Then working out $1 = \langle \alpha|\alpha\rangle = c_0\sum_n \big(|\alpha|^2\big)^n/n! = c_0 \exp\big(|\alpha|^2\big)$ gives the normalization constant $c_0.$
However note that under this Hamiltonian, $|n\rangle \mapsto e^{-in\omega t} |n\rangle$ and therefore, $$|\alpha\rangle \mapsto \exp\left(-|\alpha|^2\right) \sum_{n=0}^\infty \frac{\alpha^n}{\sqrt{n!}} ~ e^{-i~\omega t~n} |n\rangle,$$which we see on the right hand side combines by normal exponent rules to form $(\alpha e^{-i\omega t})^n.$ In other words the time evolution is that $|\alpha(t)\rangle = |\alpha_0 e^{-i\omega t}\rangle,$ and our coherent state simply makes a circle on the complex plane as it evolves.
It is in this precise sense that I understand the word "coherent," it is the meaning "it stays perfectly together as it goes along its journey." It's the same way I would say "lasers are a coherent phenomenon; light by its nature wants to spread out in a $1/r^2$ law but in a laser, the different wave packets are all arranged with just the right phase differences so that they destructively interfere for the waves that are trying to get out of the beam proper, and constructively interfere in the next position of the beam."
