a. [3 marks]

For \( H(t) = 1.63 \sin(0.513(t – 8.20)) + 2.13 \),
period = \( \frac{2\pi}{0.513} \approx 12.24 \) hours (M1).
Time from low to high tide is half the period: \( \frac{12.24}{2} = 6.12 \) hours (A1).
Convert 0.12 hours to minutes: \( 0.12 \times 60 \approx 7.2 \approx 7 \) minutes (A1).
Answer: \( m = 7 \).

b. [3 marks]

Solve \( H(t) = 1 \): \( 1.63 \sin(0.513(t – 8.20)) + 2.13 = 1 \implies \sin(0.513(t – 8.20)) = -0.693 \) (M1).
Reference angle: \( \arcsin(0.693) \approx 0.766 \) radians.
Solutions: \( 0.513(t – 8.20) = \pi + 0.766, 2\pi – 0.766 \implies t \approx 15.81, 18.96 \) (A1).
Duration: \( 18.96 – 15.81 = 3.14 \) hours (A1).
Answer: \( 3.14 \) hours.

c. [3 marks]

Differentiate \( H(t) = 1.63 \sin(0.513(t – 8.20)) + 2.13 \):
\( H'(t) = 1.63 \times 0.513 \cos(0.513(t – 8.20)) \approx 0.835 \cos(0.513(t – 8.20)) \) (M1).
At \( t = 13 \): \( 0.513(13 – 8.20) \approx 2.46 \), \( \cos(2.46) \approx -0.77 \),
so \( H'(13) \approx 0.835 \times (-0.77) \approx -0.651 \) (A1)(A1).
Answer: \( -0.651 \) m/h.

d. [4 marks]

For \( h(t) = a \sin(b(t – c)) + d \), use low tide (\( t = 2.683, h = 0.40 \)) and high tide (\( t = 9.033, h = 2.74 \)).
Solve: \( d + a = 2.74 \), \( d – a = 0.40 \implies d = 1.57 \), \( a = 1.17 \) (A1)(A1).
Time between tides: \( 9.033 – 2.683 = 6.35 \) hours, so period = \( 2 \times 6.35 = 12.7 \),
\( b = \frac{2\pi}{12.7} \approx 0.495 \) (A1).
At low tide: \( 0.40 = 1.17 \sin(0.495(2.683 – c)) + 1.57 \implies c \approx 5.86 \) (A1).
Answer: \( a = 1.17 \), \( b = 0.495 \), \( c = 5.86 \), \( d = 1.57 \).

e. [3 marks]

Solve \( H(T) = h(T) \):
\( 1.63 \sin(0.513(T – 8.20)) + 2.13 = 1.17 \sin(0.495(T – 5.86)) + 1.57 \) (M1).
Numerical solution yields \( T \approx 4.16 \) (A1)(A1).
Answer: \( T = 4.16 \).