I know standard method to compute the knot group of a trefoil $K$, regarding it as a (2,3)-torus knot. But here I find a method giving $\pi_1(\mathbb{R}^3-K)\cong\mathbb{Z}$, which is impossible. The question is that I cannot find out which step is wrong $\dots$
Here is my calculation:
Choose six points $\{A_1,\dots,A_6\} \subset K$ as showed in the picture. We can embed $K$ into $\mathbb{R}^3$ in the way such that:
(1) $A_k\in\mathbb{R}^2\times\{0\}$ for all $k$;
(2) For $k$ odd, the arcs $A_kA_{k+1}$ without endpoints lie in $\{x_3< 0\}\subset\mathbb{R}^3$;
(3) For $k$ even, the arcs $A_kA_{k+1}$ without endpoints lie in $\{x_3>0\}\subset\mathbb{R}^3$.
(We agree $A_7=A_1$)
We write $U=(\mathbb{R}^3-K)\cap\{x_3\leq 0\}$, $V=(\mathbb{R}^3-K)\cap\{x_3\geq 0\}$. Then we have:
(1) $\pi_1(U)=\pi_1(V)$ is the free group generated by 3 elements, given by loops around arcs $A_kA_{k+1}$;
(2) $U\cap V=\mathbb{R}^2-\{6pts\}$, thus $\pi_1(U\cap V)$ is the free group generated by 6 elements.
(Of course we can replace $U,V$ by a little larger open sets.)
Now we apply Van Kampen's theorem.
Let $i:U\cap V\to U$, $j:U\cap V\to V$ be inclusion, then $\pi_1(\mathbb{R}^3-K)$ is isomorphic to the free group $\pi_1(U)*\pi_1(V)$ modulo the relations $\sim$ given by $i_*(x)=j_*(x)$ for each $x\in\pi_1(U\cap V)$.
We can write $\pi_1(U)=<a_1,a_2,a_3>$, $\pi_1(V)=<b_1,b_2,b_3>$, where $a_k$ stands for a loop rotating around $A_{2k-1}A_{2k}$ and $b_k$ corresponds to $A_{2k}A_{2k+1}$. Also write $\pi_1(U\cap V)=<c_1,\dots,c_6>$, where $c_k$ corresponds to loops rotating around $A_k$. Considering orientation, we have
$$i_*(c_1)=a_1^{\pm 1}=i_*(c_2)^{-1},\ i_*(c_3)=a_2^{\pm 1}=i_*(c_4)^{-1},\ i_*(c_5)=a_3^{\pm 1}=i_*(c_6)^{-1};$$ $$j_*(c_6)=b_3^{\pm 1}=j_*(c_1)^{-1},\ j_*(c_2)=b_1^{\pm 1}=j_*(c_3)^{-1},\ j_*(c_4)=b_2^{\pm 1}=j_*(c_5)^{-1}.$$
Then $\pi_1(U)*\pi_1(V)/\sim$ $\ \ \ $ has 6 generators, each two are identical or are inverse of each other!!!
This immediately gives a group isomorphic to $\mathbb{Z}$.
So where is wrong?
Trying to explain my objection to OP's calculations (see the comments) visually.
Here's the trefoil. I made it a thin tube. Ignore its hollow interior that shows when I brutally crop the image.
Here's the top part with the 3 arches. This is a fattened version of OP's set $V$. I include the points with $x_3>-\epsilon$. Observe that this "universe" ends at the level of the bottom of the three arches.
Here's the bottom part with the 3 wormholes. This is a fattened version of OP's set $U$. I include the points with $x_3<\epsilon$. Observe that this "universe" ends at the level of the top of the three wormholes.
Here's a top-down view of the (fattened) middle part $U\cap V$ together with a base point (=the black dot) and a loop around the point $A_1$, i.e. a representative of the generator $c_1$ of $\pi_1(U\cap V)$.
Here's what the class of $c_1$ looks like in $\pi_1(U)$. We see that we can homotopically slide the loop along the wormhole $A_1A_2$. This shows, as the OP claimed, that $i_*(c_1)$ and $i_*(c_2)$ are either homotopic or homotopic to each others inverses depending on how we orient them. At least if we pick the obvious loop to represent $c_2$.
Here's how the class of $c_1$ looks like in the group $\pi_1(V)$. Observe that because in this space $c_1$ goes underneath the arch $A_4A_5$, we cannot simply slide the loop along the arch $A_1A_6$. This means that $j_*(c_1)$ is not homotopic to $b_3^{\pm}$. We need to conjugate it by $b_2$ to bring it above the arch $A_4A_5$.
The OP's attempt has other similar problems in the mappings $i_*$ and $j_*$. Working all of them out is too much work - at least for now. Another way of calculating this homotopy group (a different way of using van Kampen's theorem) is given in e.g. Massey's book.
Further observe that if choose the loop $c_1$ to go around the point $A_4$ from the West side, then the OP's claim about the image $j_*(c_1)$ disappears, but this time a similar problem appears in the bottom part, i.e. with $i_*(c_1)$.
Anyway, I'm confident that observations like this explain why OP got the wrong group to emerge, and also shows how to fix the calculations. Hope this helps in that task.