= Solution
{c}
Put $A=T^2\setminus B^2$ and let $E=\partial A=S^1$ denote the attaching <circle>. The version of the <Seifert-van Kampen theorem> used here states that if $X=U\cup V$ is an <open cover> with $U,V,U\cap V$ <path-connected> and a common <base point> in $U\cap V$, then
$$
\pi_1(X)\cong\pi_1(U)*_{\pi_1(U\cap V)}\pi_1(V).
$$
The expression on the right means the <pushout of groups> of the two inclusion-induced <group homomorphisms>; they need not be <injective>.
Choose a thin <collar neighbourhood> $C_A$ of $E$ inside $A$, and a thin equatorial band $C_S$ inside $S^2$. For example, before the square's side identifications take $C_A=\{v:1\leq|v|<1+\epsilon\}$ with $0<\epsilon<1/2$, and take $C_S=\{(x,y,z)\in S^2:|z|<\epsilon\}$. In the glued <topological space> set
$$
U=A\cup C_S,\qquad V=S^2\cup C_A.
$$
These are <open subsets>: their preimages in the two pieces are open and respect the gluing. They cover $X$, and the three <deformation retractions>
$$
U\simeq A,\qquad V\simeq S^2,\qquad U\cap V=C_A\cup_E C_S\simeq E
$$
fix $E$. Thus all three spaces are <path-connected>, and a <base point> $b\in E$ is preserved.
For completeness, the <fundamental group of a once-punctured torus> is $F(a,b)$, and its <boundary loop of a once-punctured torus> represents $[a,b]=aba^{-1}b^{-1}$, up to <orientation> and a change of <base point>. Here is the geometry behind both facts. In the square model, move each point of $A$ radially outwards to the square boundary:
$$
v\longmapsto\left((1-t)+\frac{2t}{\max(|v_1|,|v_2|)}\right)v,\qquad 0\leq t\leq1.
$$
The denominator never vanishes on $A$. The <homotopy> stays outside the missing <open disk>, stays inside the square, and fixes the square boundary, so it descends through the side identifications. The resulting <retract> is a <wedge of two circles>. Its <fundamental group> is the <free group> on the two oriented side <based loops> $a,b$. The inner <boundary loop> travels around the square once under this <deformation retraction>; its successive sides read $a,b,a^{-1},b^{-1}$.
The <sphere> $S^2$ is <simply connected>. Hence the <group homomorphism> from $\pi_1(E)\cong\mathbb Z$ to $\pi_1(V)$ is trivial, while its map to $\pi_1(U)$ sends a <group generator> to $[a,b]$. Applying the <Seifert-van Kampen theorem> gives
$$
\boxed{\pi_1(X,b)\cong F(a,b)/\langle\!\langle[a,b]\rangle\!\rangle
=\langle a,b\mid aba^{-1}b^{-1}=1\rangle\cong\mathbb Z^2.}
$$
The final <group isomorphism> follows because the single <relator> makes $a,b$ commute, leaving the <free abelian group> on two <group generators>. The two hemispheres kill the same <boundary loop>; they do not create another <group generator>.
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