All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(a)
(i)
$\frac{7 \times 6 \times 5}{7 \times 7 \times 7} = \frac{30}{49}$
Alternatively, $\frac{^7 P_3}{7^3}$

(ii)
When $n=4$, required probability $= \frac{^7 P_4}{7^3} = \frac{7 \times 6 \times 5 \times 4}{7 \times 7 \times 7 \times 7} = \frac{120}{343}$

(b)
Key GC with either sequence or use table functions.
We want to solve $\frac{^12 C_n \times n!}{12^n} \textless \frac{1}{2}$
Plot $y_1 = \frac{^12 C_x \times x!}{12^x}$ and check for the x that gives a corresponding $y_1$ value that is less than $\frac{1}{2}$

(c)
When $n =21$, probability that all 21 people have different birthdays $= \frac{^365 P_21}{365^21} = 0.55631$

Probability that all 21 people have different birthdays $= \frac{^365 P_23}{365^23} = 0.4927$

Thus, when $n =23$, probability that at least two of the people are the same birthday $= 1 - 0.4927 = 0.5073$ (more than $\frac{1}{2}$)