Why do optical phonons have such a high frequency at $k=0$? Can you give me some intuitive explanation of this? How is there any frequency when wavelength becomes almost infinite?
 A: Oscillating sublattices
Wave length (wave vector) is a measure of the phase difference between different points in space. Zero wave vector (infinite wave length) simply means that atoms in different cells oscillate in phase. As @JonCuster has correctly pointed out in the comments, optical phonons occur only in the lattices with more than one atom per unit cell. Indeed, in a monoatomic lattices talking about atoms in different unit cells oscillating in phase would not make much sense - it would mean that the whole crystal oscillates as a whole. On the other hand, in acse of multiple atoms in a unit cells we can think of optical phonons as different sublattices oscillating in respect to each other.
The Brillouin zone boundary
Let us address the extension of the question, mentioned in the comments, of why the oscillation frequency is lower at zone boundaries than at the zone center. Let us consider a lattice with two atoms, A and B, in a unit cell. Both atoms oscillate, and the distinction between the optical and acoustic phonons is essentially decomposing these oscillations into the modes where two atoms are in phase and the modes where the two atoms are in counter-phase. In the other unit cell the two atoms have the same relative phase in respect to each other, but the phases of both of them are shifted in respect to the neighboring cells. Now, there is some arbitrariness in choosing a unit cell. This is easily seen on the example of a linear lattice:
$$
-(A-B)-(A-B)-(A-B)-\\
-A-(B-A)-(B-A)-B-
$$
When the k-vector lies at the boundary of the Brillouin zone, this separation is most ambiguous - the optical and acoustic phonons are almost indistinguishable,a nd have the closest frequencies.
As a good excercise (solved in many textbooks, here is the answer) I suggest considering a chain of balls connected by springs, with alternating spring constant ($k_1$ and $k_2$) or the mass of an atom ($m_1$ and $m_2$). This can be easily solved for normal modes, resulting in acoustic and optical branches, which are split at the boundary of the Brillouin zone. However, if we take $k_1=k_2$ and $m_1=m_2$, the two branches will touch exactly at the Brillouin zone boundary!
Massive particles
What may be misleading about optical phonons is comparing them to photons: photons are massless bosons, i.e., their frequency is zero at zero wave vector. On the other hand, if we think about (relativistic) electrons, the non-zero frequency (energy) at zero momentum is not surprizing - it corresponds to the electron mass. Of course, electrosn are bosons, so it is better to compare optical phonons to bosonic massive particles - e.g., pions, which can be described by the Klein-Gordon equation. Note that this equation gives us two energy branches, which correspond to positive and negative mass solutions - the optical phonons have negative mass, i.e., their energy ahs negative curvature at $k=0$.
