(a)To find sin q =\(\frac{2\sqrt{2}}{3}\) and cos p = \(\frac{\sqrt{3}}{3}\). Then, substituting
sin(p - q) = sin p cos q - cos p sin q = (\(\frac{1}{2} \times -\frac{1}{3}\)) - (\(\frac{2\sqrt{2}}{3} \times \frac{\sqrt{3}}{2}\)
Which simplified to -\(\frac{1}{6} - \frac{2\sqrt{6}}{6}\) and can be written as \(\frac{-1 -2\sqrt{6}}{6}\)

(b), candidates were expected to obtain the following table

x	1	1.5	2	2.5	3	3.5	4
y	1	0.9428	0.8944	0.8528	0.8165	0.7845	0.7559

Then, using the table and applying the formula;
\(\int^4_1\frac{2}{\sqrt{x + 3}}\)dx = 0.5{\(\frac{1}{2}\) + 0.9428 + 0.8944 + 0.8528 + 0.8165 + 0.7845 + \(\frac{0.7559}{2}\)}
= 0.5 (5.16895)
= 2.584475
= 2.58 correct to two decimal places.