$\phi(x)|0\rangle$ is **not** the state of a particle (I stress that $\phi(x)|0\rangle$ is a *one-particle* state since I am referring to a free field) with position $x$ (when the temporal component of $x$ is zero in particular).

The situation is different form the momentum representation. Indeed, $a_p^\dagger|0\rangle$ **is** a momentum-defined one-particle state. 

The *position representation* of the particles of QFT is a quite delicate issue. It is still unclear and actually there are a number of no-go theorems about its existence either in terms of projection valued measures (PVMs) or positive-operator valued measures (POVMs).

An apparent statndard definition of the position representation is the famous [Newton-Wigner one][1], however it is plagued by a number of issues concerning locality.

A modern treatise on the issues about the position representation and the various no-go theorems, in relativistic QM can be found [here][2]
and more recently [here][3].

Probably the most powerfull no-go result is the following one. 

I should premit some facts. The *position representation* is defined by a set of operators $P_{E}$ labeled 
by sets $E \subset \mathbb{R}^3$ in the 3D rest space of a Minkowski reference space. These operators may be orthogonal projectors as in the case of a spectral measure (a PVM):
$$\vec{X} = \int_{\mathbb{R}^3} \vec{\lambda} dP(\vec{\lambda})$$  
is the triple of *position operators*. 

For instance the Newton-Wigner position operator has this form. 

A weaker formulation is the one where the $P_E$ simply define a POVM (Positive-operator valued measure) and one is dealing with the modern formulation of observables.

In both cases $\langle \psi| P_E\psi \rangle$ is the probability to find the particle in $E$ at time $t=0$. 

In the case of a POVM, every $P_E$ is simply a positive operator bounded by $I$ instead of an orthogonal projector as in the spectral decomposition where one can also take advantage of some quantum logic formulation.

One of the no-go theorems proved in [2] has the following form.

**THEOREM**(Halvorson Clifton theorem) *There is no POVM (or PVM) $P_E$, labeled by sets in the 3D space of a Minkowski mreference frame such that.*

*1) It is *covariant* under the action of spacetime translations*:
$$U^{-1}_{t,a} P_E U_{t,a} = P^{(t)}_{E-a}$$
*where $P^{(t)}$ denotes the analogous operator defined at time $t$ (the Heisemberg evolution of the initial one);*  

*2) the generator $H$ of the time translations $U(t,0)= e^{itH}$ is positive*;  
  
*3) the operators $P_E$ satisfy **locality** if $(t,E)$ and $(t',E')$ are spacelike separated then*
$$[P^{(t)}_E, P^{(t')}_{E'}]=0$$



The last requirement can be refined or weakened and it corresponds to the requirement that we cannot transmit superluminal information with the outcomes of these detectors (a version of the *no-signaling* requirment)

For instance the Newton-Wigner operator (obtained by intgrating its PVM)  violates (3) and thus cannot be considered a physical observable.

The theorem above (I stated in a rough way actually, for a precise statement see the reference I posted) is a refinement of a number of previous results due to Hegerfeldt, Borchers, Malament, Busch, in particular.

All that should answer items 3th and 4th.


Regarding the 1st and 2nd points, barring $1/\sqrt{2\omega_k}$ the only exponential  appearing in $\phi(x)|0\rangle$  is the one of $a^\dagger_k$. The phase $e^{-it \omega_k}$ (not $e^{it\omega_k}$ as it seems you wrote in 2) is the correct one since the Hamiltonian is just the factor $\omega_k$ in the momentum basis.  $\phi(t,\vec{x})|0\rangle $ is just  the Schroedinger evolution up to time $t$ of $\phi(0,\vec{x})|0\rangle$. 

Your last point (4) is quite difficult due to the presence of gauge degrees of freedom. In the Hilbert space of the photon there are only two-degrees of freedom whereas the associated quantum field has apparently 4 degrees of freedom. 

The formally added 2 degrees of freedom have their reason in making explicit the Lorentz covariance of the theory.

The one-particle state of the photon is obtained by $A_\mu(x) |0\rangle$ after removing the gauge redundancies and it can be done with several procedures (e.g. the Gupta-Bleuler one).

 However, the definition of the position operator for photons is even more difficult than for massive particles. 

  [1]:https://en.wikipedia.org/wiki/Newton%E2%80%93Wigner_localization
  [2]:https://www.worldscientific.com/doi/abs/10.1142/9789812776440_0010

  [3]: https://edoc.ub.uni-muenchen.de/25914/1/Beck_Christian.pdf