Question:

1. Given \(\cot\theta = \dfrac{7}{8}\).
\[\cot\theta = \frac{\cos\theta}{\sin\theta}=\frac{7}{8}\]
Hence \(\tan\theta = \dfrac{1}{\cot\theta}=\dfrac{8}{7}\).

2. Construct a right‑angled triangle where the side opposite \(\theta\) is 8 and the side adjacent to \(\theta\) is 7.
The hypotenuse \(h\) is
\[h = \sqrt{7^{2}+8^{2}} = \sqrt{49+64}=\sqrt{113}.\]

3. Find \(\sin\theta\) and \(\cos\theta\)
\[\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{8}{\sqrt{113}},\qquad \cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{7}{\sqrt{113}}.\]

4. Evaluate part (i)
\[\text{Expression} = \frac{(1+\sin\theta)(1+\cos\theta)}{\sin\theta-\cos\theta}.
\]
- Numerator:
\[(1+\sin\theta)(1+\cos\theta) = \left(1+\frac{8}{\sqrt{113}}\right)\left(1+\frac{7}{\sqrt{113}}\right)\]
\[= 1 + \frac{8+7}{\sqrt{113}} + \frac{8\times7}{113}
= 1 + \frac{15}{\sqrt{113}} + \frac{56}{113}.\]
- Denominator:
\[\sin\theta-\cos\theta = \frac{8-7}{\sqrt{113}} = \frac{1}{\sqrt{113}}.\]
- Division:
\[\frac{1 + \frac{15}{\sqrt{113}} + \frac{56}{113}}{\frac{1}{\sqrt{113}}}
= \left(1 + \frac{15}{\sqrt{113}} + \frac{56}{113}\right)\sqrt{113}
= \sqrt{113} + 15 + \frac{56}{\sqrt{113}}.
\]
- Combining the terms over a common denominator \(\sqrt{113}\):
\[\frac{\sqrt{113}\times\sqrt{113} + 15\sqrt{113} + 56}{\sqrt{113}}
= \frac{113 + 15\sqrt{113} + 56}{\sqrt{113}}
= \frac{169 + 15\sqrt{113}}{\sqrt{113}}.
\]
Hence
\[\boxed{\frac{(1+\sin\theta)(1+\cos\theta)}{\sin\theta-\cos\theta}=\sqrt{113}+15+\frac{56}{\sqrt{113}} = \frac{169+15\sqrt{113}}{\sqrt{113}}}.\]

5. Evaluate part (ii)
\[\cot^{2}\theta = \left(\frac{7}{8}\right)^{2}=\frac{49}{64}.\]
Hence
\[\boxed{\cot^{2}\theta = \frac{49}{64}}.\]

Final Answers
- (i) \(\displaystyle \frac{169+15\sqrt{113}}{\sqrt{113}}\) (or equivalently \(\sqrt{113}+15+\frac{56}{\sqrt{113}}\)).
- (ii) \(\displaystyle \frac{49}{64}\).