Total number of ways in which 8064 can be resolved as product of two factors is 24 as below:
(1,8064), (2,4032), (3,2688), (4,2016), (6,1344), (7,1152), (8,1008), (9,896), (12,672), (14,576), (16,504), (18,448), (21,884), (24,336), (28,288), (32,252), (36,224), (42,192), (48,168), (56,144), (63,128), (68,126), (72,112), (84,96)

One digit positive numbers = 5
Two digit positive numbers = 25
Three digit positive numbers = 100
4 digit positive numbers = 300
5 digit positive numbers = 600
Six digit positive numbers = 600
Total positive numbers,
= 5 + 25 + 100 + 300 + 600 + 600
= 1630

The letters of word 'BAKERY' be arranged in 6! ways
= 6!
= 6 × 5 × 4 × 3 × 2 × 1
= 720

Total no. of seats,
= 1 grandfar + 5 sons and daughters + 8 grandchildren
= 14
The grandchildren can occupy 4 seats on eir side of table in 4! = 24 ways.
The grandfar can occupy a seat in (5 - 1) = 4 ways (4 gaps between 5 sons and daughter).
And, remaining seats can be occupied in 5! = 120 ways (5 seat for sons and daughter).
Hence total number of required ways,
= 8! × 480

The total number of ways in which 5 part can be arranged =5! =120.
The total number of ways in which part-1 and part-3 are always toger:
= 4! × 2!
= 48
Therefore, total number of arrangements, in which y are not toger is:
= 120 - 48
= 72

1 million distinct 3 digit initials are needed.
Let number of required alphabets in language be 'n'.
Therefore, using 'n' alphabets we can form n × n × n = n³ distinct 3 digit initials.

NOTE:
Distinct initials are different from initials where digits are different. For instance, AAA and BBB are acceptable combinations in case of distinct initials while y are not permitted when digits of initials need to be different.
This n³ different initials = 1 million.
i.e. n³ = 10⁶ (1 million =10⁶)
n³ = 10²³
n = 10²
n = 100
Hence, language needs to have a minimum of 100 alphabets

Required number of ways,
= ³C₂ × 4! × 4!
= 1728

⁵C₁ × ³C₁ - 1
= 15 - 1
= 14

One among 8 gfs will get 2 gifts and remaining 7 will get one. So total of 9 gifts will be distributed among 8 gfs. i.e; 11111112
Gf who will get 2 gifts can be find out in ⁸C₁ ways = 8 ways.
Now 2 gifts can be given to selected gf in ⁹C₂ ways. And remaining 7 gifts can be given to remaining 7 gf in 7! ways.
So total no of ways= 8 × ⁹C₂ × 7!
= $$\frac{{8 \times \left( {9 \times 8} \right)}}{{2 \times 7!}}$$
= 36 × 8 × 7!
= 36 × 8!

Required number of ways
$$\eqalign{ & = {{}^4{C_2} \times {}^5{C_3}} \cr & = {{}^4{C_2} \times {}^5{C_2}} \cr & = {\frac{{4 \times 3}}{{2 \times 1}} \times \frac{{5 \times 4}}{{2 \times 1}}} \cr & = 60 \cr} $$