Fundamental group of a quotient on a solid torus. It is easy to compute the fundamental group of a solid torus. You easily get $\mathbb{Z}$ just because the torus is the cartesian product of a circumference and a closed disk.
The next step is imagine two solid torus one inside of the other one. Now consider the topological space that is the difference of the bigger one with the one that is inside. You get a torus with a hole that is a $3$-manifold with borders. Now you identify the interior border with the exterior border (in an obvious way). I worked a bit and I saw that quotient is homeomorphic to $S^1 \times S^1 \times S^1$, so the fundamental group is $\mathbb{Z}^3$.
What I am unable to compute is the fundamental group in this case that is similar to the one before. You have again bigger torus and what it is inside is homeomorphic to a torus but doing two loops inside the bigger torus (not just only one like the previous case). You do again the difference of sets and get a torus with a two loops hole inside. Now you identify borders again and you obtain a $3$-manifold with no borders. However I see myself unable to compute the fundamental group this time... 
Here goes an image of the third picture (the torus with the hole looping twice). (remember that later you identify borders). The blue tube represents the hole inside the torus.
https://www.dropbox.com/s/x2jm7gg9f9hhkw2/20150416_003300.jpg?dl=0
 A: This answer will be a bit incomplete, but I think you can work out the full details from it (feedback is welcome).
Consider you have the solid torus with the torus shaped hole winding ally around it twice, and call this space $X$, then you want to calculate $\pi_1(Z)$, where $\frac{X}{Y_1\sim_f Y_2}$, $Y_1,Y_2$ are the connected components of $\partial X$ and $f:Y_1\to Y_2$ is a homeomorphism. Now, with the notation out of the way, cover $Z$ with two open sets (Van Kampen in the making) $U,V$, where $U = \mathrm{Int} X$, and $V$ is the quotient (by the relation $\sim_f$) of an open neighborhood of $\partial X$; further, since $U\cap V$ is not connected, we are gonna have to use the groupoid version of the Van Kampen theorem.
So choose two points in $U \cap V$, which I will call $0$ and $1$ because I don't want to use more letters, and we calculate the fundamental groupoids of $U,V$ and $U\cap V$, with respect to the set of basepoints $\{0,1\}$. Each of these groupoids has two elements, and all we need to determine is the automorphism group of these two elements (corresponding to $0$ and $1$) and the morphisms between them (if they exist).


*

*For $\pi_1(U\cap V,\{0,1\})$, each element only has the fundamental group of the torus (\mathbb Z \oplus \mathbb Z) as it's automorphism group, and since $0$ and $1$ lie in different components of $U\cap V$ there are no morphisms (which correspond to paths) between them.

*For $\pi_1(V,\{0,1\})$, you can see that since $V$ deformation-retracts to the boundary of $X$ under the identification $\sim_f$. Further, you get that, once again, under this identification the boundary of $X$ is homeomorphic to the torus, and so $\hom(0,0) = \hom(1,1) = \pi_1(V) = \mathbb Z \times \mathbb Z$. Also, by choosing the base-points $0$ and $1$ such that $1 \in r^{-1}(f\circ r (0))$, one can see that all elements of $\hom(0,1)$ are of the form $\circ \gamma_{0,1} \circ k_0$, where $\gamma_{0,1}$ is a path from $0$ to $1$ and $k_0$ is an element of $\hom(0,0)$ (note that $\gamma_{0,1}\circ k_0 = h(k_0)\circ\gamma_{0,1}$ where $h$ is the "identity" $\hom(0,0) = \hom(1,1)$.

*Finally (we still have to see what are the maps, but, we're almost there) for $\pi_1(U,\{0,1\})$, we can see that $U$ deformation-retracts to a torus with a Möbius band inside it, and so has its fundamental group given by $\pi_1(X) = \frac{\mathbb Za + \mathbb Zb + \mathbb Zc}{2a + b \sim 2c}$ (there is a mistake in my comment before, sorry about that). And once again, $\hom(0,0) = \hom(1,1) = \pi_1(X)$ and $\hom(0,1)$ has elements of the form $\delta_{0,1}\circ k_0$.
Now, all that is left is to assemble the pieces by choosing a point, (either $0$ or $1$, I will choose $0$).


*

*The map $\pi_1(U\cap V,0)\to \pi_1(V,0)$ is the identitiy

*The map $\pi_1(U\cap V,0)\to \pi_1(U,0)$ takes one generator to $a$ and the other generator to $b$, as per the representation given above (note that I am assuming $0$ is the point "close to the outer boundary of $X$")

*Finally, we need to consider the group elements that arise from composing elemnts of $\hom(0,1)$ with elements of $\hom(1,0)$, these will have the form $\delta^{-1}\circ\gamma$ which is not homotopy equivalent to any of the other generators.
So, we get $\pi_1(Z) = \frac{\mathbb Za + \mathbb Zb + \mathbb Zc + \mathbb Z(\delta^{-1}\gamma)}{2a + b \sim 2c}$
A: Here's a computation using a CW-structure on $Z$. I'm not sure if it agrees or disagrres with John C. Perhaps somebody can see the error or isomorphism?
I computed the presentation $\pi_1(Z)=\left \langle a,b,c \  \middle |  \ [a,b], a[a,c], b[b,c] \right \rangle$  using the homotopic  CW complex:  http://tinyurl.com/mk94d82
First deformation retract a Mobius band bounded by the inner torus so that it becomes a circle (twice the inner torus longitude). Pick a base point on the torus. The 1-skeleton is the meridian $a$, the longitude $b$, and a curve $c$ between the inner and outer torus. Three two cells are attached: the torus giving $[a,b]=1$, a meridian cross-section giving $a[a,c]=1$, and a longitudinal cross-section giving $b[b,c]=1$. Cutting along 2-cells in $Z$ we obtain a 3-ball so that is the total cell-decomposition.
