A compound with a mass of 0.600 g contains both chlorine and bromine…

Step 1. $m(\text{AgCl}) = 0.950 - 0.376 = 0.574\ \text{g}$.

Step 2. Chlorine. $n = \dfrac{0.574\ \text{g}}{143.5\ \text{g/mol}} = 4.00\ \text{mmol}$. Mass: $4.00 \times 35.5 = 142\ \text{mg}$.

Step 3. Bromine. $n = \dfrac{0.376\ \text{g}}{188\ \text{g/mol}} = 2.00\ \text{mmol}$. Mass: $2.00 \times 80 = 160\ \text{mg}$.

Step 4. Percentages. $\%\,\text{Cl} = \dfrac{0.142\ \text{g}}{0.600\ \text{g}} \times 100 = 23.7\ \%$ and $\%\,\text{Br} = \dfrac{0.160\ \text{g}}{0.600\ \text{g}} \times 100 = 26.7\ \%$.