= Solution
Write $V=M_{S^2}(p^*/q,*)$. Filling one end of $S^1$ times an <annulus> gives a <solid torus>, so $V$ is a <solid torus>. The prescribed <algebraic intersection number of curves on an oriented surface> convention gives $\tilde f_2\cdot\tilde h_2=1$. Hence
$$
\mu_2\cdot\lambda_2=(-q^*)q-(-p)p^*=pp^*-qq^*=1.
$$
In $H_1(V;\mathbb Z)$, the first <Dehn filling> imposes $p^*f-qh_1=0$, while $h_1+h_2=0$. Therefore
$$
\boxed{\iota_{2*}(\lambda_2)=p^*f+qh_2=0.}
$$
The class $\lambda_2$ is primitive because $\gcd(p^*,q)=1$, a consequence of $pp^*-qq^*=1$. It is therefore the disk-bounding <meridian of a solid torus> of $V$. Since $\mu_2$ meets it once, <Dehn filling> $V$ along $\mu_2$ glues two <solid torus> pieces with disk boundaries meeting once. This is the standard genus-one <Heegaard splitting> of $S^3$, so
$$
\boxed{M_{S^2}(p^*/q,-q^*/p)\cong S^3.}
$$
To identify the regular fiber geometrically, put it on their common boundary <torus>. Invert the displayed basis change:
$$
\boxed{\tilde f_2=q\mu_2+p\lambda_2.}
$$
Here $\mu_2$ bounds a disk in the newly attached <solid torus> and $\lambda_2$ bounds a disk in $V$. Thus the curve has winding numbers $p$ and $q$ in the two complementary <solid torus> pieces. Both coefficients are positive and $\gcd(p,q)=1$, so it is \b[the positive $(p,q)$ torus knot]. This identifies the <torus knot> directly from the genus-one splitting, without invoking a theorem specifically about <torus knots>.
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