Solution:

I am surrounded by the enemies.

Solution:

At first, there were 2 apples on the tree. After the wind blew, one apple fell on the ground. So there where no apples on the tree and there were no apples on the ground.
*apples => more than one apple

Another solution(non tricky one)
The wind blew so hard that the apples fell of the tree and blew along the ground.

Solution:

Align the toothpicks to form TEN. Use one toothpick for each of the lines.

Solution:

Old password : altered

New Password : distinct

Solution:

30,450 miles.

Explanation:
For each 5 miles driven using this rotation, each tire will lose (2/29000)+(2/21000) of its tread.So, suppose we drive 5*N miles like this. Then, each tire will lose 2*N*((1/29000)+(1/21000)) of its tread. We wish to determine N so that the latter expression gets to be 1, representing all of each tire's tread. That happens when N=6090, which is when 5*N (the total miles driven) is 30,450.

Solution:

1653

Explanation:
For an ideal case, the batsman will hit a six on each ball. But if he hits six on the last ball of the over, the strike will change in the next over. Thus, the best he can do in the last ball is run 3 runs so that he retains the strike even in the next over. Thus the total runs that he can score in each over:
6 * 5 + 3 = 33
But this will have to go like it only till the 49th over. In the last over, he can hit a six in the last ball as well as that will be the last ball of the match.
Thus runs for the last over will be 6 * 6 = 36.
Hence the maximum runs = 33 * 49 + 36 = 1653

Solution:

Although it looks complicated, the answer is simple. The password to his computer is JIGSAW.
1st letter of Jug = J
2nd letter of Birthdays = I
3rd letter of Fights = G
4th letter of Cars =S
2nd letter of Laptops = A
1st letter of Watch = W

Solution:

Remove the same 2 matchsticks from each figure and move figure towards left 45 degrees, It forms the letters V W X Y.

Solution:

The brain teaser can be solved as shown in the image below.

Solution:

Think deeply about the door number 56, people will visit it for every divisor it has. So 56 has 1 & 56, 2 & 28, 4 & 14, 7 & 8. So on pass 1, the 1st person will open the door; pass 2, 2nd one will close it; pass 4, open; pass 7, close; pass 8, open; pass 14, close; pass 28, open; pass 56, close.
Thus we can say that the door will just end up back in its original state for each pair of divisor. But what about the cases in which the pair of divisor has analogous number for example door number 16? 16 has the divisors 1 & 16, 2 & 8, 4&4. But 4 is recurrent because 16 is a perfect square, so you will only visit door number 16, on pass 1, 2, 4, 8 and 16… leaving it open at the end. So only perfect square doors will remain open at the end.

Solution:

Selfie
Explanation:
: Selfie = is six letter word
: Self = The first four letter is me.
: e = The second and last letter are the same.
: fee = The fourth second and last letter are payment.