(1a) is correct.
A constant function has a single value in its range, and there are $25$ possible values to choose from.
(1b) is wrong.
As Calvin pointed out in the comments, you did not impose the required condition in your counting. Note that if $25$ is in the range of $f$, then $25$ is automatically the maximum value of $f$. Hence, to count functions of this form, it suffices to count functions where $25$ appears at least once among the images of $f$. Equivalently, one can exclude the functions that don't have $25$ in their range. So the answer is $25^{25}-24^{25}$.
(1c) is wrong.
Though you did recognize that $f(1) \neq f(2) \neq f(3)$ (they must take on distinct values), your counting does not necessarily force $f(1) < f(2) < f(3)$. For example, suppose you choose $f(1) = 25$. Then you do not have $24$ options for $f(2)$. In fact, you have no options that satisfy $f(2) > f(1)$. So you have overcounted.
To remedy this, note that you don't have to assign $f(1), f(2)$ and $f(3)$ directly (why does your counting do this?). Instead, if you simply choose $3$ numbers for $f(1), f(2)$ and $f(3)$ to take on, there is a unique order that makes $f(1) < f(2) < f(3)$. So the problem is equivalent to choosing $3$ numbers for $f(1), f(2)$ and $f(3)$, then mapping the remaining $22$ inputs to any of the $25$ possible values. Hence, the answer is ${25 \choose 3}25^{22}$.
(2a) is wrong.
The idea of using complements is correct, but was executed incorrectly. Let $M_1, M_2$ be our distinguished men. The total committees of size $k$ is ${n+m \choose k}$. What you have considered your total, ${n \choose k}{m \choose k}$, on the other hand, is counting something else - say, committees of size $2k$ with $k$ men and $k$ women. As for counting the committees that violate the condition (where $M_1, M_2$ are in them), this is equivalent to fixing $M_1, M_2$ in the committee, then choosing the remaining $k-2$ other people from the other $n+m-2$ individuals. So ${n+m-2 \choose k-2}$ committees violate the condition. Instead, you counted ${n-2 \choose k}$, which counts, say, the committees of size $k$ comprised only of men excluding $M_1$ and $M_2$. The correct count gives us an answer of ${n+m \choose k } - {n+m-2 \choose k-2}$.
(2b) is wrong.
The answer is the same as (2a), since the argument directly transfers. It does not matter that they are of different gender. You are still excluding a specific pair of individuals.
(3a) is correct.
This is a case of "stars and bars".
(3b) is wrong.
One approach would be to give $2$ boards to each school up front, then distribute the rest freely. If we give $2$ boards to each school up front, that leaves us with $35-2(4) = 27$ boards to distribute freely, reducing to the case of part (a) to get an answer of ${ 27+4-1 \choose 4-1} = {30 \choose 3}$.
(4) is unclear. What is $r$?
To solve this, one can try to reduce it to the case of counting when a vector of nonnegative integers sums to a specific number, for which the technique is just to use stars and bars. To do this, we modify our problem a bit. Since $x_i > 0$, we have $x_i -1 \geq 0$.
Let $y_i = x_i - 1$. If $x_1 + \cdots + x_n \leq n+2$, then
\begin{align} y_1 + \cdots + y_n &= (x_1 - 1) + \cdots + (x_n - 1) \\&= x_1 + \cdots + x_n - n \\&\leq (n+2)-n = 2. \end{align}
This looks very similar to the problem we want to reduce to. Indeed, the only difference is that we have a $\leq$ instead of an equals (note that the $y_i$ are nonnegative). To fix this, observe that if $$y_1 + \cdots + y_n \leq 2,$$ then there is some $z \geq 0$ such that $$y_1 + \cdots + y_n + z = 2.$$ So a solution $(x_1, \dots, x_n)$ to the original problem corresponds directly to a solution $(y_1,\dots,y_n,z)$ of nonnegative integers such that $$y_1 + \cdots + y_n + z = 2,$$ and vice versa (let $x_i = y_i+1$). This completes our reframing of the problem into the one linked above. By stars and bars, the number of solutions $(y_1, \dots, y_n, z)$ of this form is precisely ${2+(n+1)-1 \choose (n+1)-1} = {n+2 \choose n} = {n+2 \choose 2}$.
