= Solution
Use the region labels and curve orientations from part (a). A <Dehn presentation of a knot group> generator goes from the basepoint above the projection plane through its region and returns through the unbounded region. Equivalently it is dual to the correspondingly labeled $\alpha$ disk of the <Heegaard diagram>. A crossing <relator> records the signed intersections of its $\beta$ curve with these disks. Setting $o=1$ gives
$$
\boxed{\pi_1(X_8)=\langle d,e,a,b,c\mid ae^{-1}d^{-1},\ ad^{-1}cb^{-1},\ ba^{-1}ec^{-1},\ cd^{-1}e^{-1}\rangle.}
$$
For example, the overpassing strand at $A$ equates the <meridian of a knot> words $d$ and $ae^{-1}$; at $B$ it equates $ad^{-1}$ and $bc^{-1}$. The other two crossings equate $ba^{-1}$ with $ce^{-1}$ and $cd^{-1}$ with $e$.
Choose the region-generator direction so that $e$ is the shown <meridian of a knot> $\mu$. The <Alexander numbering> of the regions, starting from zero outside and increasing by one on crossing an oriented strand from right to left, is
$$
\boxed{\varphi(d)=\varphi(e)=\varphi(\mu),\qquad \varphi(a)=\varphi(c)=2\varphi(\mu),\qquad \varphi(b)=3\varphi(\mu).}
$$
One can also obtain this directly by <abelianization> of the <relators>: $a=d+e$, $c=d+e$, $b=a-d+c$, and $b-a+e-c=0$. The last two imply $d=e$, and the resulting <first homology group> is freely generated by $e$. Thus $H_1(X_8;\mathbb Z)\cong\mathbb Z$, with no extra torsion.
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