Two dice are rolled. List all possible Sample spaces. Taka adding of each possiblities which are get the highest odd?

Each die has 6 faces (1, 2, 3, 4, 5, 6). When two dice are rolled, the sample space $S$ consists of all ordered pairs $(d_1, d_2)$, where $d_1$ is the result of the first die and $d_2$ is the result of the second die.

$$S = \begin{Bmatrix} (1,1) & (1,2) & (1,3) & (1,4) & (1,5) & (1,6) \\ (2,1) & (2,2) & (2,3) & (2,4) & (2,5) & (2,6) \\ (3,1) & (3,2) & (3,3) & (3,4) & (3,5) & (3,6) \\ (4,1) & (4,2) & (4,3) & (4,4) & (4,5) & (4,6) \\ (5,1) & (5,2) & (5,3) & (5,4) & (5,5) & (5,6) \\ (6,1) & (6,2) & (6,3) & (6,4) & (6,5) & (6,6) \end{Bmatrix}$$

There are $6 \times 6 = 36$ possible outcomes in the sample space.

The total for each possibility is the sum of the numbers on the two dice ($d_1 + d_2$).
The minimum sum is $1+1=2$.
The maximum sum is $6+6=12$.
The possibility that yields the highest total is when both dice show the highest possible number.
The highest total is obtained from the outcome $(6,6)$, which sums to $6+6=12$.

The possibility with the highest total is $\boxed{(6,6)}$.

2) If three dice are tossed, what is the probability of obtaining an even number on all three?

Each die has 6 possible outcomes (1, 2, 3, 4, 5, 6).
For three dice, the total number of possible outcomes is $6 \times 6 \times 6$.

$$\text{Total outcomes} = 6^3 = 216$$

For a single die, the even numbers are {2, 4, 6}. There are 3 even outcomes.
For three dice to show an even number on all three, each die must land on an even number.

$$\text{Favorable outcomes} = 3 \times 3 \times 3 = 27$$

The probability of an event is given by the ratio of favorable outcomes to the total number of possible outcomes.

$$P(\text{all three even}) = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{27}{216}$$

$$P(\text{all three even}) = \frac{27}{216} = \frac{1}{8}$$

The probability of obtaining an even number on all three dice is $\boxed{\frac{1}{8}}$.