1. You should definitely be proving it, there's no reason for the pattern to continue. However, in this case, it's sort of obvious why $\mathbb{P}(X \geq n) = p^{n-1) + q^{n-1}$ for $n\geq 3$. I would justify by saying that there can't have been a mummy and daddy penguin amongst the first $n-1$ boxes so it follows, but avoid saying "because the pattern continues". I'd suspect that for this particular question, you wouldn't lose any marks for just writing down $p^{n-1} + q^{n-1}$ because of how straightforward it is, but I'd caution against this in general. Justify your answers.

2. Your approach is fine. Other ways include:

(i) $pq = p(1-p) = \frac{1}{4} - \left(p-\frac{1}{2}\right)^2 \leq \frac{1}{4}$ so $\frac{1}{pq} \geq 4.$
(ii) Use the AM-GM inequality that says that $\sqrt{pq} \leq \frac{p+q}{2}$ but $p+q = 1$ so $pq \leq 4$ so $\frac{1}{pq} \geq 4$
(iii) Write $\frac{1}{pq} = \frac{(p+q)^2}{pq}$ and then work with that.

(iv) there are loads of methods, but yours is fine tbh. (although I would personally justify that you don't need to check the endpoints (p=0 or p=1) because $p,q \neq 0$ from the STEM, although you wouldn't lose any marks, extra justification won't hurt - remember that the derivative only tells you that there's a minimum in (0,1) [p=0 or p=1 might be global minimums])