Why is $d*F$ equal to $\partial _\mu F^{\mu \nu}$? Given that $A = A_\nu dx^\nu$ and $F = \partial_{\mu}A_\nu dx^\mu \wedge dx^\nu$
Why does $d*F$ equal to $\partial _\mu F^{\mu \nu}$?
How does all the $\frac{1}{2}\varepsilon^{abcd}F_{cd}$ fit into this picture, and how do you compute this stuff explicitly? It's so confusing...
 A: In 4 dimensions and using the coordinates $(t,x,y,z)$ on $\mathbb R^4$ one can write
$$ 
F :=\frac{1}{2}F_{\mu\nu} dx^{\mu} \wedge dx^{\nu}= \\
B_x dy \wedge dz + B_y dz \wedge dx + B_z dx \wedge dy + E_x dx \wedge dt + E_y dy \wedge dt + E_z dz \wedge dt$$
which is equal to to $F =dA = ( \partial_{\mu}  A_{\nu} ) dx^{\mu} \wedge dx^{\nu}$.
We need to apply the Hodge star operator * on the basis $(dt, dx, dy, dz)$ of $\Omega^1(\mathbb R^4)$, i.e.
$${\star dy} \wedge dz = - dx \wedge dt ,\quad {\star dz} \wedge dx = -dy \wedge dt,\quad{\star dx} \wedge dy = - dz \wedge dt ,\\
{\star dx} \wedge dt = dy \wedge dz,\quad {\star dy} \wedge dt = dz \wedge dx,\quad
{\star dz} \wedge dt = dx \wedge dy, $$ 
(in 4 dimensions applying the Hodge star operator to a 2-form one arrives at a 4-2=2-form) obtaining
$$\star F = - B_x dx \wedge dt - B_y dy \wedge dt - B_z dz \wedge dt + E_x dy \wedge dz + E_y dz \wedge dx + E_z dx \wedge dy$$


*

*How to finish the proof
Now we need to apply the exterior derivative to each term of $*F$. We arrive at the sum of 12 terms which can be grouped into 6 3-forms; we finish by proving that  $d*F=0$ is equivalent to $\partial_\mu F^{\mu\nu}=0$, for all $\nu=1,\dots,4$.
A caveat has to be said here: $d*F=0$ is an equality in $\Omega^3(\mathbb R^4)$ where, by construction, $d*F$ is the linear combination of the 4 linear independent 3-forms
$$dt\wedge dx\wedge dy,~dt\wedge dx\wedge dz,~dx\wedge dy\wedge dz,~dt\wedge dy\wedge dz $$
which are a basis of $\Omega^3(\mathbb R^4)$.
The equality $d*F=0$ is then satisfied if and only if the coefficients of these 4 3-forms are all equal to $0$ by linear independence. We are left to prove  that this last condition is equivalent to the scalar equations $\partial_\mu F^{\mu\nu}=0$.
We apply the exterior derivative operator $d$ to $*F$; all we need is to properly group all terms; we obtain (shortly)
$$d*F = (\partial_y B_x -\partial_xB_y +\partial_t E_z)dt\wedge dx\wedge dy +
(-\partial_z B_y +\partial_yB_z +\partial_t E_x)dt\wedge dy\wedge dz+\dots
 $$
and similarly for the remaining contributions. A quick check shows that the coefficient of $dt\wedge dx\wedge dy$ is the sum
$$-\partial_\mu F^{\mu4},$$
while the coefficient of $dt\wedge dy\wedge dz$ is equal to 
$$\partial_\mu F^{\mu2}.$$
The remaining cases are similar.
EDIT
This more recent answer is related to the OP and contains additional computations
A: As an addition to the answer by Avitus, I would like to give you some hints how such calculations could be tackled in general. Since a lot of signs and combinatorial factors are involved I give all results up to a multiplicative constant, this, however does not affect the result. 
I denote the anti-symmetrization by $A_{[\alpha}B_{\beta]}:=\frac{1}{2}(A_\alpha B_\beta-A_\beta B_\alpha)$ etc.
We get:
$$
\begin{align}
F &= F_{\mu\nu}dx^\mu\wedge dx^\nu = F_{[\mu\nu]}dx^\mu\wedge dx^\nu\\
\Rightarrow\star F&\propto F^{[\mu\nu]}\epsilon_{\mu\nu\rho\sigma}\;dx^\rho\wedge dx^\sigma\\
\Rightarrow d\star F&\propto\partial_\lambda F^{[\mu\nu]}\epsilon_{\mu\nu\rho\sigma}\;dx^\lambda\wedge dx^\rho\wedge dx^\sigma\\
&\propto \partial_\lambda F^{[\mu\nu]}\delta_{[\mu}^\lambda\epsilon_{\nu]\sigma_1\sigma_2\sigma_3}dx^{\sigma_1}\wedge dx^{\sigma_2}\wedge dx^{\sigma_3}\qquad(\dagger)\\
&\propto \partial_{[\mu} F^{[\mu\nu]}\epsilon_{\nu]\sigma_1\sigma_2\sigma_3}dx^{\sigma_1}\wedge dx^{\sigma_2}\wedge dx^{\sigma_3}\\
&=\partial_{\mu} F^{[\mu\nu]}\epsilon_{\nu\sigma_1\sigma_2\sigma_3}dx^{\sigma_1}\wedge dx^{\sigma_2}\wedge dx^{\sigma_3}\\
&\overset{!}{=}0\quad\Rightarrow\quad\partial_{\mu} F^{[\mu\nu]}=\partial_{\mu} F^{\mu\nu}=0\quad\forall \nu
\end{align}
$$
To get $(\dagger)$ try to use that for any tensor $K$ in $d$ dimensions
$$
K^{[i_1\dots i_k]}\propto\epsilon^{i_1\dots i_kj_{k+1}\dots j_d}\epsilon_{l_1\dots l_kj_{k+1}\dots j_d}K^{l_1\dots l_k}
$$
to show that
$$
\epsilon_{\mu_1\dots\mu_k\nu_{k+1}\dots\nu_d}K^{[\lambda_1\dots\lambda_l\nu_{k+1}\dots\nu_d]}\propto\delta_{[\mu_1\dots}^{\lambda_1\dots}\delta_{\mu_l}^{\lambda_l}(\star K)_{\mu_{l+1}\dots\mu_k]}
$$
Then take $K^{[\lambda\rho\sigma]}=dx^\lambda\wedge dx^\rho\wedge dx^\sigma$ and compute using the above formula
$$
\epsilon_{\mu\nu\rho\sigma}K^{[\lambda\rho\sigma]}
$$
In total it's just a mess with taking hodge duals ($\star K\sim\epsilon^{\dots}K$) and anti-symmetrization ($K^{[\dots]}\sim\epsilon^{\dots}\epsilon_{\dots}K$). I would at most remember that $\star\sim\epsilon$ and "anti-symm"$\sim\epsilon^2$
A: This is easier using "geometric calculus," the calculus built upon clifford algebra.
There, instead of $d$ and $\star d$, we use $\partial \wedge$ and $\partial \cdot$ as notations.  So the divergence of the Faraday bivector is just denoted
$$\partial_a F^{ab} = \partial \cdot F$$
Hodge duality is performed by multiplication with the pseudoscalar $\epsilon$.  So what we're being asked to compute is
$$d (\star F) = \partial \wedge (F \epsilon)$$
Identity:  $\partial \wedge (F \epsilon) = (\partial \cdot F)\epsilon$.  The proof uses "grade projection", in which we use the associativity of the clifford algebra's geometric product.
$$\langle \partial F \epsilon \rangle_3 = \langle (\partial F) \epsilon \rangle_3 = \langle \partial (F \epsilon) \rangle_3$$
These middle term is just $(\partial \cdot F) \epsilon$ and the last term is $\partial \wedge (F\epsilon)$.  This is exhaustive, so the identity is proven.
