Solution:

The task was simple as you can see in the picture. This is how you will cut and join the boards and then you will be able to cover the hole completely.

Solution:

This puzzle has no answer if you are considering this figure in a 2d domain. However if you follow it on a three-dimensional plane, you can do it like shown in the figure given.

Solution:

I + IX = X

The equation read from the other side becomes X = XI + I which is false.

Solution:

one fish, one cat and one dog.

Explanation:
The question might appear a bit difficult in the starting but if you analyze the statements, you will realize that it is just a tricky one.

All except two were fishes and all except two were cats. With these statements we can assume that two of the animals were not fishes and two were not cats. Now one of those animals that are not fishes can be a cat and one of those two animals that are not cats can be a fish. Just carry out the same analysis for the statement that all except two animals were not dogs and you will come across the result i.e.:
In that competition, there was one fish, one cat and one dog.

Solution:

For this answer is 3^0, 3^1, 3^2... That is 1,3,9,27,81,243 and 729.

Solution:

2 degree below zero

Solution:

7, 4, 6, 5, 8

Solution:

To do it, the marbles must be arranged as the vertices of an octahedron. If you are unsure, check the image below.

Solution:

4 6 5 3 (from magician perspective)
3 5 6 4 (viewer perspective)

Detail:

From user perspective:
3 5 6 4 => 5 6 4 3 => 6 5 4 3

For magician perspective
3 4 5 6 (ascending order as)

Solution:

Put one black rock in one sack and put all the rest in the other one. In such manner you will get about 3/4 probability of surviving.

Solution:

Numbers of the first cube:
1, 2, 4, 0, 5 and 6.

Numbers on the second cube:
3, 1, 2, 7, 8 and 0.

Now you can pick any date and you will be able to show it using these cubes. Remember that you can always use 6 as 9 by inverting it.

This problem has another solution and we leave it for you to deduce.