Far side of the moon question

Moons are not always tidally locked.

They are formed with a certain spin angular momentum that will be dragged by asymmetries on the hosting planetary body. Those asymmetries then backreact onto the moon and drag it's spin to become synchronized with the orbit angular momentum.
How rapidly this process happens is roughly a function of the tidal forces
$$F_{Tid} = G frac{m_{Moon} cdot m_{Planet}}{r^3} d$$
exerted by those asymmetries. Here $d$ is the moon size, $r$ the distance between planet and moon and the other quantities are masses and the gravitational constant.

The time $t_{lock}$ this locking takes can be estimated by comparing the excess angular momentum $L_{ex} = r^2m_{Moon}(omega_{actual} - omega_{orbit})$ with those acting forces:
$$t_{lock} approx L_{ex} / left(r F_{Tid} right) approx frac{r^5}{ d cdot m_{Planet}}$$

The latter equation makes it fairly obvious that this is not a process that happens at a certain magical orbit or at a certain time. It is an asymptotic process, and the only coincidal thing is that we observe any given moon at a time where it may not have lost enough spin angular momentum yet.
This should hopefully also clarify some confusion from the comments:

• Yes, the tidal locking time is also dependent on the mass of the host-planet, but much more sensitive to the planet-moon-distance. Note that this derivation is independent of the moon's mass, and only dependent on it's size. The literature knows much more sophisticated variants of this, that also incorporate a (realistic) weak dependency on the moon's mass.


• For Hot Jupiter exoplanets it is thought they'd be in tidal lock when the asteroseismologic age of the system is bigger than the derived tidal lock time.

Concluding
1.) Any moon would be tidally locked if we could wait long enough.
2.) There are much more effects / torques to consider for a detailed understanding than I did here.