The answer is -
(ways of selecting $3$ out of $5$ from I and $5$ out of $7$ from II) $+$
(ways of selecting $4$ out of $5$ from I and $4$ out of $7$ from II) $+$
(ways of selecting $5$ out of $5$ from I and $3$ out of $7$ from II)
$$
(5 \mathrm{C} 3) \times(7 \mathrm{C} 5)+(5 \mathrm{C} 4) \times(7 \mathrm{C} 4)+(5 \mathrm{C} 5) \times(7 \mathrm{C} 3)=420
$$
NOTE that the following answer/approach would be WRONG:
Find the mistake in doing it in this way:
(ways of selecting the minimum $3$ from each part I and II and $2$ out of the remaining $6$ from both parts)
$$
(5 \mathrm{C} 3) \times(7 \mathrm{C} 3) \times(6 \mathrm{C} 2)=5250.
$$
This Mistake is The Most Common Mistake that students make.
The mistake is that it's over-counting.
The second method would count selecting questions $1,2,3$ from part I, $1,2,3$ from part II, and then $4$ and $5$ from part I again, multiple times. Look if we rearrange how we choose:
$1,4,5$ from part I, $1,2,3$ from part II, and $2,3$ from part I also.
The second method would consider these two as two different combinations while clearly, they are the same.
