(a) This can either be done using the long division method (easier) or convert everything to base ten and divide, then re-convert to base two.
I'll be using the latter method.
\(11111111_{two} = 1 \times 2^{7} + 1 \times 2^{6} + 1 \times 2^{5} + 1 \times 2^{4} + 1 \times 2^{3} + 1 \times 2^{2} + 1 \times 2^{1} + 1 \times 2^{0}\)
= \(128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 255_{ten}\)
\(101_{two} = 1 \times 2^{2} + 0 \times 2^{1} + 1 \times 2^{0} \)
= \(4 + 0 + 1 = 5_{ten}\)
\(\frac{255}{5} = 51_{ten}\)

2	51
2	25 r 1
2	12 r 1
2	6 r 0
2	3 r 0
2	1 r 1
	0 r 1

= \(110011_{two}\)
(b)
(i) Length of arc AXB = \(\frac{105}{360} \times 2 \times \frac{22}{7} \times 6cm\)
= \(11cm\)
\(\therefore Perimeter = 6cm + 11cm + 6cm = 23cm\)
(ii) Area of sector = \(\frac{105}{360} \times \frac{22}{7} \times 6cm \times 6cm\)
= \(33 cm^{2}\)