a) Using the principle of multiplication, the answer is $6^4 = 1296$.

b) Firstly, there are 4 ways to choose $i$ such that $x_i$ is equal to the sum of the remaining numbers. For each $i$, we choose $x_i$ first and using the star-bar problem, one can obtain $\binom{x_i-1}{2}$ ways to choose the other 3 numbers. For $x_i = 3, 4, 5, 6$; we find that there are
$$\binom{2}{2} + \binom{3}{2} + \binom{4}{2} + \binom{5}{2} = 1 + 3 + 6 + 10 = 20$$
tuples $(x_1, x_2, x_3, x_4)$ such that $x_i$ is equal to the sum of the remaining numbers. Thus, there are $4 \times 20 = 80$ tuples satisfying the condition and the probability is $\frac{80}{1296} = \frac{5}{81}$.

c) Denote $S_1, S_2, S_3$ to be the sets of tuples that satisfy
$$x_1 + x_2 = x_3 + x_4, \quad x_1 + x_3 = x_2 + x_4, \quad x_1 + x_4 = x_2 + x_3.$$
We consider the case $x_1 + x_2 = x_3 + x_4$, which is equivalent to
$$(x_1 - 1) + (x_2 - 1) + (6 - x_3) + (6 - x_4) = 10.$$
Let $y_1 = x_1 - 1$, $y_2 = x_2 - 1$, $y_3 = 6 - x_3$, $y_4 = 6 - x_4$ then
$$\begin{cases} 0 \le y_1, y_2, y_3, y_4 \le 5, \\ y_1 + y_2 + y_3 + y_4 = 10. \end{cases} \quad (1)$$
Denote $A$ to be the set of tuples $(y_1, y_2, y_3, y_4)$ such that
$$\begin{cases} y_1, y_2, y_3, y_4 \ge 0, \\ y_1 + y_2 + y_3 + y_4 = 10. \end{cases} \quad (2)$$
By using star-bar problem, we have
$$|A| = \binom{13}{10} = 286.$$
Denote $A_i$ to be the set of tuples $(y_1, y_2, y_3, y_4)$ that satisfy (2) and $y_i \ge 6$. Assume that $y_1 \ge 6$ or
$$\begin{cases} y_1 \ge 6, y_2, y_3, y_4 \ge 0, \\ (y_1 - 6) + y_2 + y_3 + y_4 = 6. \end{cases}$$
Using the star-bar problem, we obtain $|A_1| = \binom{7}{4} = 35$. Similarly, we get $|A_2| = |A_3| = |A_4| = 35$. It is also clear that $A_1, A_2, A_3, A_4$ are disjoint. The number of tuples $(y_1, y_2, y_3, y_4)$ that satisfy (1) is
$$|A| - (|A_1| + |A_2| + |A_3| + |A_4|) = 146.$$
Hence, the number of tuples $(x_1, x_2, x_3, x_4)$ that $x_1+x_2 = x_3+x_4$ is 146 or $|S_1| = 146$.
Clearly, $S_1 \cap S_2$ is the set of tuples $(x_1, x_2, x_3, x_4)$ such that
$$\begin{cases} x_1 + x_2 = x_3 + x_4, \\ x_1 + x_3 = x_2 + x_4, \end{cases} \quad \longleftrightarrow \quad \begin{cases} x_1 = x_4, \\ x_2 = x_3. \end{cases}$$
Thus, there are 6 ways to choose $x_1$ and $x_4$, 6 ways to choose $x_2$ and $x_3$, which implies $|S_1 \cap S_2| = 6^2 = 36$. Similarly,
$$|S_2 \cap S_3| = |S_3 \cap S_1| = 36.$$
In case $S_1 \cap S_2 \cap S_3$, one can check that $x_1 = x_2 = x_3 = x_4$. There are 6 ways to choose $x_1, x_2, x_3, x_4$ which implies that $|S_1 \cap S_2 \cap S_3| = 6$. By applying the principle of inclusion and exclusion, we obtain
$$\begin{aligned} |S| &= |S_1 \cup S_2 \cup S_3| \\ &= (|S_1| + |S_2| + |S_3|) - (|S_1 \cap S_2| + |S_2 \cap S_3| + |S_3 \cap S_1|) \\ &\quad + |S_1 \cap S_2 \cap S_3| = 146 \cdot 3 - 36 \cdot 3 + 6 = 336. \end{aligned}$$
Combining with part b), the number of tuples $(x_1, x_2, x_3, x_4)$ such that we can divide the numbers in two groups that have the same sum which is $336 + 80 = 416$ or the probability is $\frac{416}{1296} = \frac{26}{81}$.
$\square$