You can divide your task in steps, count the ways that you can conduct each step and then use the multiplication principle to count the total number of ways that you can conduct your task.
a) You can do it in two steps.
1.Step. Order your books. You can do it in $25!$ ways.
2.Step. Decide on how many books will be placed on each shelf. That is equivalent to writing on piece of paper 10 nonnegative (zeros are here allowed in c they are not!) integers $x_1,x_2,\dots,x_{10}$ so that they add up to 25. For $i=1,2,\ldots,10$, $x_i$ represents the number of books that will be placed on the $i$-th shelf. The total number of ways that you can conduct this step is equal to the number of solutions to the equation $$x_1+x_2+\ldots+x_{10}=25,$$ with $0\leq x_i, i=1,2,\ldots,10$. This is a typical stars and bars problem with $n=10$ and $k=25$. Then (according to the Wikipedia page, Stars and Bars, Theorem two) you can do that in $$\dbinom{n+k-1}{n-1}=\dbinom{10+25-1}{10-1}=\dbinom{34}{9}$$
So in sum the task in (a) can be conducted in $25!\times\dbinom{34}{9}$ ways.
c). Based on a) the answer is straightforward by allowing in Step 2. only positive values for the $x_i, i=1,2,\ldots,10$. The total number of ways that you can conduct this step is equal to the number of solutions to the equation $$x_1+x_2+\ldots+x_{10}=25,$$ with $1\leq x_i, i=1,2,\ldots,10$. Then (according to the Wikipedia page, Stars and Bars, Theorem one) you can do it in $\dbinom{k-1}{n-1}=\dbinom{25-1}{10-1}$ ways.
So in sum the task in (c) can be conducted in $25!\times\dbinom{24}{9}$ ways.
b) 25C25 is equal to 1 so, your approach here is not correct. Here you can think it as follows. Write on each book a number from 1 to 10 (this number indicates the shelf that it will be placed). So you can do it $10\times10\times\ldots\times10=10^{25}$ ways. (With steps as in the previous questions: 1.Step Choose a shelf for book Nr1. You can do it in 10 ways. 2.Step Choose a shelf for book Nr2. You can do it in 10 ways and so on until 25.Step Choose a shelf for book Nr25. You can do it in 10 ways. So by multiplication rule you can do it in $10^{25}$ ways.)
Let $A,B,C$ be considered a single element$=M$ (say)
So we have four elements-$M,D,E,F$
These $4$ elements can be arranged in $4!$ ways.
And the $3$ elements in $M$ can be arranged in $3!$ ways.
So the required number of ways = $4! \times 3! = 24 \times 6 = 144$
Best Answer
Your textbook answer seems off too:
All together: $$7! \cdot 3! = 30240$$