Solution:

Give Up

SLOW when read down gives us Slow Down.
GIVE when read upwards gives us Give Up.

Solution:

There is no reason whatever why the customer's original deposit of Rs.100 should equal the total of the balances left after each withdrawal.
The total of withdrawals in the left-hand colum may equal Rs.100, but is is purely coincidence that the total of the right-hand column is close to Rs.100.
Let us show another example, but starting with Rs.200 in the bank:
Withdrawals Balance left
Rs.50 Rs.150
Rs.25 Rs.125
Rs.10 Rs.115
Rs.8 Rs.107
Rs.5 Rs.102
Rs.2 Rs.100
Rs.100 Rs.699
Moral of this story? Don't Total Balances

Solution:

He askes the god to beat him half to death.

Solution:

Everyone will done one day or other.

Abraham Lincoln era is too old and we expect most of them(if not all) to be dead by now.

Solution:

73

Explanation:
It's a very famous puzzle known as "Josephus Problem Puzzle". It can be solved in various ways but the trick to solve this problem is illustrated below.

Binay number of 100 is: 1100100
complement pf 100 is 0011011, whose decimal equivalent is 27.
The desired answer will be 100 - 27 = 73

Solution:

10

Explanation:
Equation 1 + 9 + 11 = 1 can be derived from
One (o) + nine (n) + eleven (e) = one => 1

Similarly for equation,
12 + 11 + 9
Twelve (t) + eleven (e) + nine (n) => ten (10)

Solution:

If you see the picture, they formed a pentagonal shape and achieved the required.

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:

At each draw, the number of strawberry candies are either decreasing by 2 or not decreasing at all. In the case of orange candies, at each draw, they are either increasing by 1 or decreasing by 1.

Thus on an assumed outset with at least one candy in the jar to begin with, if the number of strawberry candies are 0 or are even in numbers, they will finish off leaving an orange candy at the end. If otherwise, the remaining candy will be a strawberry one.

Solution:

Yes it is possible in Roman Numerals. In Roman Numerals,
9 = IX
10 = X
11 = XI

Thus by removing one from 9, you are getting 10 and by removing one from 11, you are getting 10 again.

Solution:

In the picture you can find the green matchsticks. We have moved them from the bottom and placed them in the middle. It was a bit tricky but now you have your desired result.