a) To find the value of \( m \):
The water height at Sule Skerry is \( H(t) = 1.63 \sin(0.513 \times (t – 8.20)) + 2.13 \).
Period: \( \frac{2\pi}{0.513} \approx 12.238 \) hours.
Time from low to high tide is half the period:
\( \frac{12.238}{2} \approx 6.119 \) hours.

Convert to hours and minutes:
\( 6.119 – 6 = 0.119 \) hours.
\( 0.119 \times 60 \approx 7.14 \) minutes.

Rounding to the nearest integer:
\( m \approx 7 \) [3]

b) To find the time when the water height is less than 1 metre between consecutive high tides:
Solve \( H(t) < 1 \):
\( 1.63 \sin(0.513 \times (t – 8.20)) + 2.13 < 1 \)
\( 1.63 \sin(0.513 \times (t – 8.20)) < -1.13 \)
\( \sin(0.513 \times (t – 8.20)) < -\frac{1.13}{1.63} \approx -0.693 \)

Find boundary points where \( H(t) = 1 \):
\( \sin(0.513 \times (t – 8.20)) = -0.693 \)
Reference angle: \( \arcsin(0.693) \approx 0.766 \) radians.
Solutions in third and fourth quadrants:
\( 0.513 \times (t – 8.20) = \pi + 0.766 + 2k\pi \quad \text{or} \quad 2\pi – 0.766 + 2k\pi \)
\( = 3.908 + 2k\pi \quad \text{or} \quad 5.518 + 2k\pi \)

For one cycle starting at the first high tide (\( t \approx 11.261 \)) to the next (\( t \approx 23.499 \)):
For \( 0.513 \times (t – 8.20) = 3.908 \):
\( t – 8.20 = \frac{3.908}{0.513} \approx 7.616 \)
\( t \approx 8.20 + 7.616 = 15.816 \) hours.

For \( 0.513 \times (t – 8.20) = 5.518 \):
\( t – 8.20 = \frac{5.518}{0.513} \approx 10.756 \)
\( t \approx 8.20 + 10.756 = 18.956 \) hours.

Duration where \( H(t) < 1 \):
\( 18.956 – 15.816 \approx 3.14 \) hours.

Thus:
The length of time is approximately 3.14 hours [5]

c) To find the rate of change of water height at \( t = 13 \):
Compute the derivative:
\( H(t) = 1.63 \sin(0.513 \times (t – 8.20)) + 2.13 \)
\( H'(t) = 1.63 \times 0.513 \times \cos(0.513 \times (t – 8.20)) \approx 0.83619 \cos(0.513 \times (t – 8.20)) \)

Evaluate at \( t = 13 \):
\( H'(13) = 0.83619 \cos(0.513 \times (13 – 8.20)) \)
\( 13 – 8.20 = 4.80 \)
\( 0.513 \times 4.80 \approx 2.4624 \)
\( \cos(2.4624) \approx -0.778 \)
\( H'(13) \approx 0.83619 \times (-0.778) \approx -0.651 \)

Thus:
The rate of change is approximately \( -0.651 \) m/h [3]

d) To find the values of \( a, b, c, d \):
For Rockall: \( h(t) = a \sin(b \times (t – c)) + d \).
Given:
– Low tide at \( t = 2.6833 \) (02:41), height 0.40 m.
– High tide at \( t = 9.0333 \) (09:02), height 2.74 m.

Determine \( a \) and \( d \):
\( a = \frac{2.74 – 0.40}{2} = 1.17 \)
\( d = \frac{2.74 + 0.40}{2} = 1.57 \)

Low tide at \( t = 2.6833 \):
\( 0.40 = 1.17 \sin(b \times (2.6833 – c)) + 1.57 \)
\( 1.17 \sin(b \times (2.6833 – c)) = -1.17 \)
\( \sin(b \times (2.6833 – c)) = -1 \)
\( b \times (2.6833 – c) = -\frac{\pi}{2} + 2k\pi \quad (1) \)

High tide at \( t = 9.0333 \):
\( 2.74 = 1.17 \sin(b \times (9.0333 – c)) + 1.57 \)
\( 1.17 \sin(b \times (9.0333 – c)) = 1.17 \)
\( \sin(b \times (9.0333 – c)) = 1 \)
\( b \times (9.0333 – c) = \frac{\pi}{2} + 2m\pi \quad (2) \)

For first occurrences (\( k = 0 \), \( m = 0 \)):
\( b \times (2.6833 – c) = -\frac{\pi}{2} \quad (3) \)
\( b \times (9.0333 – c) = \frac{\pi}{2} \quad (4) \)

Subtract (3) from (4):
\( b \times (9.0333 – c – (2.6833 – c)) = \frac{\pi}{2} – \left(-\frac{\pi}{2}\right) \)
\( b \times 6.35 = \pi \)
\( b \approx \frac{\pi}{6.35} \approx 0.495 \)

Substitute \( b \approx 0.495 \) into (3):
\( 0.495 \times (2.6833 – c) = -\frac{\pi}{2} \approx -1.5708 \)
\( 2.6833 – c \approx \frac{-1.5708}{0.495} \approx -3.177 \)
\( c \approx 2.6833 + 3.177 \approx 5.860 \)

Thus:
\( a = 1.17 \), \( b \approx 0.495 \), \( c \approx 5.863 \), \( d = 1.57 \) [5]

e) To find \( T \) where \( H(T) = h(T) \):
\( 1.63 \sin(0.513 \times (T – 8.20)) + 2.13 = 1.17 \sin(0.495 \times (T – 5.863)) + 1.57 \)

Simplify:
\( 1.63 \sin(0.513 \times (T – 8.20)) – 1.17 \sin(0.495 \times (T – 5.863)) = -0.56 \)

Define:
\( f(T) = 1.63 \sin(0.513 \times (T – 8.20)) – 1.17 \sin(0.495 \times (T – 5.863)) + 0.56 \)
Solve \( f(T) = 0 \) numerically, testing around Rockall’s low tide (\( t = 2.6833 \)):

At \( T = 4.16 \):
\( H(4.16) \approx 1.63 \sin(0.513 \times (4.16 – 8.20)) + 2.13 \approx 0.60 \)
\( h(4.16) \approx 1.17 \sin(0.495 \times (4.16 – 5.863)) + 1.57 \approx 0.60 \)

Thus:
\( T \approx 4.16 \) [3]