Write down the value of \( a \).

The function is \( f(x) = a \cos(bx) \). From the graph, the y-intercept is at \( (0, 5) \).

Evaluate at \( x = 0 \):

\[ f(0) = a \cos(b \cdot 0) = a \cos(0) = a \cdot 1 = 5 \]

\[ a = 5 \]

Answer: \( a = 5 \).

(i) Write down the period of \( f \).

The period is the length of the interval for one complete cycle of the function. From the graph, one full cycle occurs from \( x = 0 \) to \( x = \pi \).

\[ \text{Period} = \pi \]

(ii) Find the value of \( b \).

For \( f(x) = a \cos(bx) \), the period is given by \( \frac{2\pi}{|b|} \).

Given the period is \( \pi \):

\[ \frac{2\pi}{|b|} = \pi \]

\[ \frac{2}{|b|} = 1 \implies |b| = 2 \]

Since \( b \in \mathbb{Z}^+ \), \( b = 2 \).

Answer: (i) Period = \( \pi \), (ii) \( b = 2 \).

Find the value of \( f\left(\frac{\pi}{6}\right) \).

From parts (a) and (b), \( a = 5 \), \( b = 2 \), so \( f(x) = 5 \cos(2x) \).

Evaluate at \( x = \frac{\pi}{6} \):

\[ f\left(\frac{\pi}{6}\right) = 5 \cos\left(2 \cdot \frac{\pi}{6}\right) = 5 \cos\left(\frac{\pi}{3}\right) \]

\[ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \]

\[ f\left(\frac{\pi}{6}\right) = 5 \cdot \frac{1}{2} = \frac{5}{2} \]

Answer: \( f\left(\frac{\pi}{6}\right) = \frac{5}{2} \).