It is given that \(\cos A=-\frac{60}{61}\) and \(\tan B=\frac{-7}{24}\) and neither \(A\) nor \(B\) is in the second quadrant, then \(A\) and \(B\) will be in third and fourth quadrant respectively. So, \(\frac{B}{2}\) will be in second quadrant. Now,
\(\begin{aligned}
\tan B & =-\frac{7}{24} \\
\Rightarrow \quad \frac{2 \tan \frac{B}{2}}{1-\tan ^2 \frac{B}{2}} & =-\frac{7}{24}
\end{aligned}\)
\(\begin{aligned} & \Rightarrow \quad 7 \tan ^2 \frac{B}{2}-48 \tan \frac{B}{2}-7=0 \\ & \Rightarrow\left(7 \tan \frac{B}{2}+1\right)\left(\tan \frac{B}{2}-7\right)=0 \\ & \Rightarrow \quad \tan \frac{B}{2}=-\frac{1}{7}\left[\because \frac{B}{2} \in \text { IInd quadrant }\right] \\ & \because \tan \left(A+\frac{B}{2}\right)=\frac{\tan A+\tan \frac{B}{2}}{1-\tan A \tan \frac{B}{2}}=\frac{\frac{11}{60}-\frac{1}{7}}{1+\frac{11}{420}} \\ & {\left[\because \cos A=-\frac{60}{61} \text { and } A \in \text { IIIrd }\right.} \\ & \left.\text { Quadrant } \therefore \tan A=\frac{11}{60}\right]\end{aligned}\)
\(=\frac{77-60}{420+11}=\frac{17}{431}\)
\(\because A \in\) IIIrd quadrant and \(\frac{B}{2} \in\left(\frac{3 \pi}{4}, \pi\right)\) and \(\tan \left(A+\frac{B}{2}\right)\) is positive.
\(\therefore\left(A+\frac{B}{2}\right) \in\) Ist quadrant.
Hence, option(a) is correct.