1) Your answer is correct; for each ball, you can choose any box, and every choice is distinguishable at any time.
2) You want to distribute your 5 distinguishable balls into 3 indistinguishable boxes. Let $B(5,3)$ denote the number of ways in which this can be done into exactly 3 indistinguishable non-empty boxes, and use the recurrence relation $B(n,k)=B(n-1,k-1)+kB(n-1,k)$ with $B(n,1)=1$ and $B(n,n)=1$. You seek $B(5,1)+B(5,2)+B(5,3)$. For further reference, see Stirling number.
3) You want to separate your 5 indistinguishable balls into 3 distinguishable boxes. Note that every admissible separation uniquely corresponds to a permutation of the string $XXbbbbb$.
4) You want to partition your 5 indistinguishable balls into 3 indistinguishable boxes. Let $p(5,3)$ denote the number of ways in which this can be done into exactly 3 indistinguishable non-empty boxes, and use the recurrence relation $p(n,k)=p(n-1,k-1)+p(n-k,k)$ with $p(n,1)=1$ and $p(n,n)=1$. You seek $p(5,1)+p(5,2)+p(5,3)$. For further reference, see partition.
First problem:
How many ways are there to distribute $26$ identical balls into six distinct boxes such that the number of balls in each box is odd?
Let $x_k$ denote the number of balls placed in the $k$th box, $1 \leq k \leq 6$. Then
$$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 26 \tag{1}$$
Since each $x_k$ is a positive odd integer, $x_k = 2y_k + 1$ for some non-negative integer $y_k$. Substituting into equation 1 and simplifying yields
$$y_1 + y_2 + y_3 + y_4 + y_5 + y_6 = 10 \tag{2}$$
which is an equation in the non-negative integers. The number of solutions of equation 2 is the number of ways five addition signs can be inserted into a row of ten ones.
There are $$\binom{10 + 5}{5} = \binom{15}{5}$$ solutions since we must choose which five of the fifteen symbols (ten ones and five addition signs) will be addition signs.
Second problem:
How many ways are there to distribute $26$ identical balls into six distinct boxes such that the first three boxes contain at most six balls?
We wish to solve the equation
$$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 26 \tag{3}$$
in the non-negative integers subject to the constraints that $x_1, x_2, x_3 \leq 6$. Equation 3 is an equation in the non-negative integers. It has
$$\binom{26 + 5}{5} = \binom{31}{5}$$
solutions.
We must exclude those cases in which one or more of the constraints is violated.
Suppose the constraint $x_1 \leq 6$ is violated. Then $x_1 \geq 7$. Let $y_1 = x_1 - 7$. Then $y_1$ is a non-negative integer. Substituting $y_1 + 7$ for $x_1$ in equation 3 and simplifying yields
$$y_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 19 \tag{4}$$
Equation 4 is an equation in the non-negative integers. It has
$$\binom{19 + 5}{5} = \binom{24}{5}$$
solutions. By symmetry, equation 3 has the same number of solutions if the constraint $x_2 \leq 6$ or the constraint $x_3 \leq 6$ is violated. Hence, there are
$$\binom{3}{1}\binom{24}{5}$$
solutions in which one of the constraints is violated.
Suppose the constraints $x_1 \leq 6$ and $x_2 \leq 6$ are both violated. Let $y_1 = x_1 - 7$; let $y_2 = x_2 - 7$. Then $y_1$ and $y_2$ are non-negative integers. Substituting $y_1 + 7$ for $x_1$ and $y_2 + 7$ for $x_2$ and simplifying yields
$$y_1 + y_2 + x_3 + x_4 + x_5 + x_6 = 12 \tag{5}$$
which is an equation in the non-negative integers with
$$\binom{12 + 5}{5} = \binom{17}{5}$$
solutions. By symmetry, equation 3 has the same number of solutions if both the constraints $x_1 \leq 6$ and $x_3 \leq 6$ are violated or both the constraints $x_2 \leq 6$ or $x_3 \leq 6$ are violated. Hence, there are
$$\binom{3}{2}\binom{17}{5}$$
solutions in which two of the constraints are violated.
Suppose that all three constraints are violated. Let $y_k = x_k - 7$, $1 \leq k \leq 3$. Then each $y_k$ is a non-negative integer. Substituting $y_k + 7$ for $x_k$, $1 \leq k \leq 3$, and simplifying yields
$$y_1 + y_2 + y_3 + x_4 + x_5 + x_6 = 5 \tag{6}$$
which is an equation in the non-negative integers with
$$\binom{5 + 5}{5} = \binom{10}{5}$$
solutions.
By the Inclusion-Exclusion Principle, the number of ways the $26$ identical balls can be distributed into six distinct boxes such that the first three boxes contain at most six balls is
$$\binom{31}{5} - \binom{3}{1}\binom{24}{5} + \binom{3}{2}\binom{17}{5} - \binom{3}{3}\binom{10}{5}$$
Best Answer
I'll solve the case in which the balls are identical. This is not the nicest solution, but it works and I hope it is clear.
Case 1: First two boxes have exactly four balls.
If the first two boxes together contain four balls, then we can use stars and bars twice. The number of ways to place four balls in two boxes is the number of ways to arrange $4$ stars and $1$ bar, or $5$ choose $1$. The number of ways to place the remaining four balls in the other four boxes is the number of was to arrange $4$ stars and $3$ bars, or $7$ choose $3$. This gives
$$\binom{5}{1}\binom{7}{3}$$
Case 2: First two boxes have exactly three balls. With the same reasoning, we get
$$\binom{4}{1}\binom{8}{3}$$
Case 3: First two boxes have exactly two balls.
$$\binom{3}{1} \binom{9}{3}$$
Case 4: First two boxes have exactly one ball.
$$\binom{2}{1}\binom{10}{3}$$
Case 5: First two boxes have no balls.
$$\binom{1}{1}\binom{11}{3}$$
(This makes sense because this is just stars and bars applied to putting $8$ balls into $4$ boxes)
Our total is therefore
$$5\binom{7}{3}+4\binom{8}{3}+3\binom{9}{3}+2\binom{10}{3}+\binom{11}{3}=\boxed{1056}$$