The position collapses into a position eigenstate (a spike in wavefunction) because you measured the position. That spike spreads over time because the position eigenstates are not momentum eigenstates, i.e. there is a range of momenta in the spike. Another way of looking at this is that the spike is not an eigenstate of the Hamiltonian (i.e. a well defined energy) and therefore evolves in time beyond simple a phase change.
If you measure the energy, you will collapse into an energy eigenstate, which is not a position eigenstate. For example, energy eigenstates of a particle in a box (classic quantum homework problem) are distributed across the width of the box, not uniformly, but not strongly localized like the spike. If you measure the energy again, you will get the same answer (it is an eigenstate of the Hamiltonian).
As for measurements, we measure energy differences usually. For example, the energy eigenstates of a hydrogen atom (or hydrogen like atom) correspond to various configurations of the electron "orbiting" the atom plus relative orientation of the electron's magnetic moment (spin) to the "orbit" and the nuclear magnetic moment.
I work with calcium ions that effective have a single electron around a nucleus (the other electrons effectively cancel out). If the calcium ion is in the lowest energy configuration (ground state) and I shine a laser on it, sweeping the frequency of the laser, the ion will remain dark until my laser is in a narrow ~20 MHz window around 755 THz (397 nm UV light) at which point the ion will glow ... something we can see with a sensitive CCD camera. That 755 THz corresponds to the energy difference between two different configuration of the ion. The ion glows because it can absorb the light at that specific frequency and will then re-emit the light into a random direction through spontaneous emission. The absorption and emission happen about 20 million times a second. If the ion is not in the ground state, and there are other metastables levels it could be in, the ion won't glow. In this way I can detect whether it was in the ground state. When I turn off the laser, the ion will emit a single final photon and fall back into the ground state (and you can detect this photon).
If the ion was originally in some superposition of the ground state and one of the metastable states, the ion will either collapse into the ground state and glow or collapse into the metastable state and stay dark. The probability of it being bright or dark depends on the original superposition. Through this measurement process, I have collapsed the ion's state into an energy eigenstate. At no point are we measuring the electron's position or momentum, only the energy difference.
The following link about last year's Nobel prize might help.
There are two answers to this question. The first is that, yes, in non-relativistic quantum mechanics, you can have things going faster than the speed of light, because relativity is never taken into account. The fix is to learn quantum field theory.
We can also consider this particular situation more closely. Your thought experiment suggests that a momentum measurement can "teleport" a particle infinitely far away in an infinitely short time, which feels unphysical, regardless of relativity.
The solution is that a precise momentum measurement takes a finite amount of time. (And an infinitely precise momentum measurement, as you're suggesting, takes infinitely long.)
To see this, start from the energy-time uncertainty principle
$$\Delta E \Delta t \geq \hbar$$
where $\Delta E = \Delta (p^2/2m) = p \Delta p / m$. Then we have the bound
$$\frac{p \Delta p \Delta t}{m} \geq \hbar$$
where $p$ is the (average) value of momentum you get, $\Delta p$ is the uncertainty on that momentum, and $\Delta t$ is the time it took to perform the measurement. This tells us that more precise momentum measurements take longer.
Now, the final state after this 'smeared' momentum measurement is a wavepacket centered on the origin with width $\Delta x$, with
$$ \Delta x \Delta p \sim \hbar.$$
Finally, combining this with our other result gives
$$ \frac{\Delta x}{\Delta t} \leq \frac{p}{m}.$$
That is, the particle is not moving any faster than it would be, semiclassically.
Best Answer
No, it doesn't collapse to an eigenstate. Collapse to an eigenstate is a picture of an ideal measurement. In general the final state will not be describable by a wave function, because it's not a pure state, it is instead a mixed state. See this question, which is about inexact measurements.
Position eigenstate in position representation is $\langle x_{}|x_0\rangle=\delta(x-x_0)$. This gives the following in the momentum representation: $\langle p_{}|x_0\rangle=e^{\frac{i}{\hbar}px}$. For this function probability density is constant, thus its expectation value is undefined (one can't find a center of infinite line). Similarly, for free particle expectation value of energy will also be undefined. This is because such state is an abstraction, a useful mathematical tool. Of course, such states can't be prepared in real experiment, but one can come very close to it, e.g. shoot an electron at a tiny slit and observe state of the electron at the very exit of that slit.
As to finding expectation value of energy in position eigenstate, first mistake which you make using the formula $\overline E=\langle x|\hat H|x\rangle$ is forgetting to normalize the eigenvector. But position operator has continuous spectrum, which makes all its eigenvectors unnormalizable (i.e. if you try to normalize them, you'll get null vector, which is meaningless as a state). Thus you can't directly find expectation value of energy in position eigenstate.