(N/A) Total cards in a deck = $52$.
Number of jacks = $4$,queens = $4$,kings = $4$. Total removed = $4 + 4 + 4 = 12$.
Remaining cards $n(S) = 52 - 12 = 40$.
$(i)$ Let $E_1$ be the event of getting a card with value $7$. There are $4$ such cards (one of each suit).
$P(E_1) = \frac{n(E_1)}{n(S)} = \frac{4}{40} = \frac{1}{10}$.
$(ii)$ Let $E_2$ be the event of getting a card with a value greater than $7$. The possible values are $8, 9, 10$. There are $3$ values,each with $4$ suits,so $n(E_2) = 3 \times 4 = 12$.
$P(E_2) = \frac{n(E_2)}{n(S)} = \frac{12}{40} = \frac{3}{10}$.
$(iii)$ Let $E_3$ be the event of getting a card with a value less than $7$. The possible values are $1, 2, 3, 4, 5, 6$. There are $6$ values,each with $4$ suits,so $n(E_3) = 6 \times 4 = 24$.
$P(E_3) = \frac{n(E_3)}{n(S)} = \frac{24}{40} = \frac{3}{5}$.