To add the given numbers $(9F.3F)_{16}$ and $(277.36)_{8}$, we first convert both numbers into binary (base-2) form. First, convert $(9F.3F)_{16}$ to binary: Each hexadecimal digit corresponds to a 4-bit binary sequence. $$ \begin{aligned} 9_{16} &= 1001_2 \\ F_{16} &= 1111_2 \\ 3_{16} &= 0011_2 \\ F_{16} &= 1111_2 \end{aligned} $$ Thus, $(9F.3F)_{16} = (10011111.00111111)_2$. Second, convert $(277.36)_{8}$ to binary: Each octal digit corresponds to a 3-bit binary sequence. $$ \begin{aligned} 2_8 &= 010_2 \\ 7_8 &= 111_2 \\ 7_8 &= 111_2 \\ 3_8 &= 011_2 \\ 6_8 &= 110_2 \end{aligned} $$ Thus, $(277.36)_8 = (010111111.011110)_2 = (10111111.01111000)_2$. Now, add the two binary numbers: $$ \begin{aligned} & 010011111.00111111_2 \\ + & 010111111.01111000_2 \\ \hline & 101011110.10110111_2 \end{aligned} $$ Therefore, the sum is $(101011110.10110111)_2$.