$A, B, C$ are angles of a triangle.
$$
\begin{aligned}
& \therefore A+B+C=180^{\circ} \Rightarrow A+B=180^{\circ}-C \\
& \Rightarrow \quad \frac{A+B}{2}=90^{\circ}-\frac{C}{2}
\end{aligned}
$$
Now, $\cos A+\cos B+\cos C$
$$
\begin{aligned}
& =2 \cos \left(\frac{A+B}{2}\right) \cdot \cos \left(\frac{A-B}{2}\right)+\left(1-2 \sin ^2 \frac{C}{2}\right) \\
& =2 \cos \left(90^{\circ}-\frac{C}{2}\right) \cdot \cos \left(\frac{A-B}{2}\right)+1-2 \sin ^2 \frac{C}{2} \\
& =2 \sin \frac{C}{2} \cdot \cos \left(\frac{A-B}{2}\right)-2 \sin ^2 \frac{C}{2}+1 \\
& =2 \sin \frac{C}{2}\left[\cos \left(\frac{A-B}{2}\right)-\sin \frac{C}{2}\right]+1 \\
& =2 \sin \frac{C}{2}\left[\cos \left(\frac{A-B}{2}\right)-\cos \left(\frac{A+B}{2}\right)\right]+1 \\
& =2 \sin \frac{C}{2}\left[2 \sin \frac{A}{2} \cdot \sin \frac{B}{2}\right]+1 \\
& \Rightarrow \cos A+\cos B+\cos C=1+4 \sin \frac{A}{2} \sin \frac{B}{2} \cdot \sin \frac{C}{2}
\end{aligned}
$$
On comparing, we get
$$
\begin{aligned}
& a=1, b=4 \\
\therefore \quad & a+b=5
\end{aligned}
$$