= Solution
Let $x,y$ be the generators associated with the two pairs of <one-handle> feet. Following the attaching circle across the identifications marked $a,b$ and $c,d$ gives the word $xyx^{-1}y^{-1}$, up to reversal and changing generator orientations. Thus the <van Kampen theorem> gives
$$
\boxed{\pi_1(W)=\langle x,y\mid xyx^{-1}y^{-1}=1\rangle\cong\mathbb Z^2.}
$$
The <handle decomposition> has cellular groups $C_0=\mathbb Z$, $C_1=\mathbb Z^2$, $C_2=\mathbb Z$, with no higher groups. The boundary of the two-cell is the abelianized attaching word, which is zero. Hence
$$
\boxed{H_j(W;\mathbb Z)=\begin{cases}\mathbb Z,&j=0,2,\\\mathbb Z^2,&j=1,\\0,&j\ge3.\end{cases}}
$$
The zero framing is the product framing of the <commutator handlebody of the torus>, so the represented manifold is $T^2\times D^2$ and its <intersection form> is $(0)$. In particular its boundary is $T^2\times S^1=T^3$.
The boundary <homology (mathematics)> can also be obtained without using the product identification. By <Poincare-Lefschetz duality>, $H_3(W,\partial W)=H^1(W)=\mathbb Z^2$ and $H_2(W,\partial W)=H^2(W)=\mathbb Z$. The map $H_2(W)\to H_2(W,\partial W)$ is its zero <intersection form>. The <long exact sequence of a pair> then gives split sequences
$$
0\to\mathbb Z^2\to H_2(\partial W)\to\mathbb Z\to0,\qquad
0\to\mathbb Z\to H_1(\partial W)\to\mathbb Z^2\to0.
$$
The boundary is connected and oriented, so its zeroth and third <homology (mathematics)> are both $\mathbb Z$. Therefore
$$
\boxed{H_j(\partial W;\mathbb Z)=\begin{cases}\mathbb Z,&j=0,3,\\\mathbb Z^3,&j=1,2,\\0,&j\ge4.\end{cases}}
$$
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