Q1: yes, it does.
Q2: "Electromagnetic fields are the same as waves"
Not always, e.m. fields may be static - static electric field around charges and static magnetic fields around magnets or (DC) currents, or waves - e.g as emitted by an antenna.
"so does that mean that when current is induced, they get the energy from electromagnetic waves being exchanged, or do they induce current directly from the electromagnetic field?"
In your example you don't have waves, you have a static magnetic field.
Q3: "If electromagnetic fields and waves are the same,"
As I said, they are not always the same - see my answer to question Q2.
"doesn't that mean that you can induce current from waves? Therefore transferring electricity over long distances?"
You can send waves over long distances, see how works the antennas, the lasers, the communication through satellites (don't forget that light is also e.m. field).
Q4: "What spectrum of light are the electromagnetic waves that are in a field created by electricity?"
This question is not clear - what you mean "e.m. waves that are in a field created by electricity"? A field around static electric charges contains no waves. Maybe after you read my answers above, you'll be able to express more clearly what you ask.
Q5: "Why is it not possible to induce current from the Earth's magnetic field? I thought Maxwell said that electric fields and magnetic fields were the same thing."
The law of induction of electromotive force is $\mathscr E = -\text d \Phi / dt$ where $\Phi$ is the magnetic flux. Thus, for generating an electromotive force $\mathscr E$ a variable magnetic field is needed in your coil. I am not sure whether the magnetic field of the Earth varies in time at all, and surely not as quickly and in the form as needed for producing current in your coil. Anyway, it's not known to me that we generate electricity from the Earth's magnetic field, see in Wikipedia how we generate current from magnetic flux.
Note that for an oscillating dipole:
$$ {\bf p}({\bf r},t)= {\bf p}({\bf r})e^{-i\omega t}$$
the fields are:
$${\bf E}=\frac 1{4\pi\epsilon_0}\big[\frac{\omega^2}{c^2r}(\hat{\bf r}\times{\bf p})\times\hat{\bf r}+\big(\frac 1 {r^3}-\frac{i\omega}{cr^2}\big)(3\hat{\bf r}[\hat{\bf r}\cdot{\bf p}]-{\bf p}) \big]e^{i \omega r/c}\ \ e^{-i \omega t} $$
and
$${\bf B}=\frac{\omega^2}{4\pi\epsilon_0\,c^3}(\hat{\bf r}\times {\bf p})(1-\frac c{i\omega r}) \frac{e^{i \omega r/c}}r\ e^{-i \omega t} $$
which shows that there is a lot going on.
Note that there are different powers of $r$ in different term, which means that the field configuration looks different at different distances.
Many wavelengths away (the far-field, or radiation zone), only the smallest inverse powers of $r$ term contribute. It is here that the $E$ and $B$ fields are nice sine waves with their peaks in phase.
At smaller $r$ is the transition region, and the smallest $r$--typically within a few wavelengths--is the near field. Note that these terms pick up factors of $i$, which means the $E$ and $B$ fields can be out of phase, which is what you would expect from simple induction.
The near field depends not only on the antenna design, but also on conductors that may be in the near field. For critical applications (say a spacecraft lander leg in the near field of your terminal descent radar), you can't just imagine what is going on, and detailed calculations or measurements may be required.
The dipole formulae shown here are the simplest analytical case; however they do illustrate the origin of the differences in near are far field configurations.
Best Answer
You write the integral formulation of Faraday's law, but there is also the equivalent differential formalism: $$ \nabla\times\mathbf E=-\frac{\partial\mathbf B}{\partial t}\tag{1} $$ which can be proven in a straight-forward manner (and ought to be done in your standard E&M textbooks).
Using standard planar waves equations, $$ E_y=E_0\sin\left(kx-\omega t\right) \\ B_z=B_0\sin\left(kx-\omega t\right), \\ $$ then Equation (1) gives us that $$ kE_0\cos(kx-\omega t)=\omega B_0\cos(kx-\omega t) $$ Which enforces the well-known relation that $E_0=cB_0$. So clearly plane waves do satisfy Faraday's law.
A similar situation exists with Ampere's law, usually written as $$ \oint\mathbf B\cdot d\mathbf l=\frac{1}{c^2}\frac{d\phi_E}{dt} $$ which leads to $$ \nabla\times\mathbf B=\frac{1}{c^2}\frac{\partial\mathbf E}{\partial t} $$ (I'm ignoring the current density here).