Compute the fundamental and homology groups of $S^3 \setminus K$, where $K$ is two linked copies of $S^1$ in $\mathbb R^3$ 
Compute the homology groups of $S^3 \setminus K$, where $K$ is two
  linked copies of circles in $\mathbb R^3$.  How about the homology
  group of $S^3 \setminus K'$ where $K'$ is just one copies of
  circles in $\mathbb R^3$?

What is really important and interesting is how to compute the homology group. To compute $H_*(S^3 \setminus K)$ I guess we should use Mayer-Vietoris sequence  to $A$ and $B$, two copies of $S^3\setminus S^1$.
However, I even don's know how to compute $H_*(S^3 \setminus S^1)$. To do this, I try Mayer-Vietoris theorem again.
Because I have few intuition about $S^3$, I do not know what the maps are in there Mayer-Vietoris long exact sequences.
Please help!
 A: For basic Mayer-Vietoris usage you need to thicken your link/knot a little in such way that you won't create new intersections in obtained in such way "link" of solid tori.
Namely: let $K$ be your link of $n$ circles and $U_1 = K \times D^2_{3 \varepsilon}$, where $D_{3 \varepsilon}^2$ is disk of "small" radius $3 \varepsilon$ such that you can retract $U_1$ to $K$. Then let's denote by $U_2 = S^3 \setminus \{K \times D^2_{ \varepsilon}\}$ it's thickened complement. MV sequence gives us:
$$\ldots\rightarrow H_2(U_1 \cup U_2) \rightarrow H_1(U_1 \cap U_2) \rightarrow H_1(U_1) \oplus H_1(U_2) \rightarrow  H_1(U_1 \cup U_2) \rightarrow \ldots$$
$$\ldots\rightarrow H_2(S^3) \rightarrow H_1(U_1 \cap U_2) \rightarrow H_1(U_1) \oplus H_1(U_2) \rightarrow  H_1(S^3) \rightarrow \ldots$$
$$\ldots0 \rightarrow H_1(U_1 \cap U_2) \rightarrow H_1(U_1) \oplus H_1(U_2) \rightarrow  0 \rightarrow \ldots.$$
Observe now that:


*

*$U_1 \cap U_2$ is homologically disjoint sum of $n$ tori and hence $H_1(U_1 \cap U_2) = \bigoplus_{i=1}^n \mathbb{Z}\oplus \mathbb{Z}$

*$H_1(U_1) = \bigoplus_{i=1}^n \mathbb{Z}$

*$H_1(U_2) = H_1(S^3 \setminus K)$
hence $H_1(S^3 \setminus K) = \bigoplus_{i=1}^n \mathbb{Z}$.
The problem with this approach is that it depends on the embedding of $K$ in $S^3$. It may not be possible to find described small $\varepsilon$ (search for "wild knots"). However Qiaochu already mentioned the workaround - Alexander Duality which gives the result immediately.
