Fundamental group, homology of the complement of the "figure eight" knot in $S^3$? The "figure eight" knot is the embedded circle in the three-sphere. What is the fundamental group and the homology of the complement of the knot in $S^3$?
 A: This is a standard example in Rolfsen, which I suggest you find a copy of if you are going to look further into these kinds of questions.  Here is a link to what he says.
First, the figure 8 knot is also called $4_1$ (the first knot with 4 crossings).  On page 56 and 57 he describes the algorithm for writing down the Wirtinger Presentation of a knot diagram.  And on 58 he gives the fundamental group of Euclidean space minus the knot, $\pi_1(\mathbb{R}^3-4_1)$, which is the same as $\pi_1(S^3-4_1)$, which answers your first question.
Then for the homology:  It is well known the $H_1$ is the abelianization of $\pi_1$. So you will notice that when you let everything in $\pi_1(S^3-4_1)$ commute, you get just $\mathbb{Z}$. This is true for every knot.  
Again, I will redirect you to Rolfsen, page 50 for the other homology groups.  It happen $H_*(S^3-K)$ are are all isomorphic to  $H_*(S^3-S^1)\cong H_*(S^2)$.  Intuitively, you should think about homology as not being a fine enough invariant to capture knotting.  It only sees an embedded circle as the standardly embedded unknot.  
