Answer: (a) \(35.6\text{ cm}\); (b) \(68.0\text{ cm}^2\).

(a) The perimeter of the shaded region is made from \(AC\), \(CB\), and the arc \(AB\).

Since \(OB=16\) and \(OC=7.5\),

\(CB=16-7.5=8.5.\)

Use the cosine rule in triangle \(AOC\):

\(AC^2=16^2+7.5^2-2(16)(7.5)\cos\frac{2\pi}{7}.\)

The arc length \(AB\) is

\(16\cdot\frac{2\pi}{7}.\)

Therefore the perimeter is

This gives

\(35.6\text{ cm}\)

to 3 significant figures.

(b) The shaded area is the area of sector \(AOB\) minus the area of triangle \(AOC\).

The sector area is

\(\frac12(16^2)\frac{2\pi}{7}.\)

The triangle area is

\(\frac12(16)(7.5)\sin\frac{2\pi}{7}.\)

So

This gives

\(68.0\text{ cm}^2\)

to 3 significant figures.