 Solution

Let $X_{\#}$ be the connected-sum exterior and fill it along its Seifert longitudeThe decomposing annulus in $X_{\#}$ has two parallel boundary circles, each meeting ${\lambda }_{\#}$ once. Applying part (b) cuts the filling back into $X_1$ and $X_2$ and identifies the peripheral curves byup to the signs required by the orientation-reversing torus gluing.

In $Y_1^{\prime }=X_1({\lambda }_1)$, the core $K_1^{\prime }$ has meridian ${\mu }_1^{\prime }={\lambda }_1$. By the orientation stipulated in the question its longitude is ${\lambda }_1^{\prime }={\mu }_1$. Thus the displayed gluing sends the meridian and longitude of $K_1^{\prime }$ to the longitude and meridian of $K_2$, respectively, which is exactly the splice of knots. Therefore the required integer surgery is zero surgery on the connected sum: