You just need a few algebraic properties, perhaps you remember that multiplying two negative numbers produces a positive number $(-2)\cdot(-3)=6$. When you multiply this with another negative number the result becomes negative again. $(-2)\cdot(-3)\cdot(-5)=6\cdot(-5)=-30$. If you look at multiplications with even more negative factors you can see a pattern emerge, that whenever the number of negative factors is odd the result will be negative and positive when the number of negative factors is even.
Then you need to recall, that the power notation $a^n$ is just an abrreviation of a long product $a^n=\underbrace{a\cdot a\cdot a\cdot \ldots\cdot a}_{n\text{ times}}$. Perhaps you also know that when you raise a product to a power you can distribute the power, like so $(a\cdot b)^n=a^ n\cdot b^n$.
The last bit you need to answer the question is, that you always can factor out any term you might need. For example in the expression $(3-2x)$ there is no number $5$ in it. But if I want to have a $5$, or any other number in it, I just have to use $\frac{5}{5}=1$. Because multiplying with $1$ doesn't change anything. So I can put in almost any place I like $(3-2x)=\left(\frac{5}{5}3-\frac{5}{5}2x\right)=5\left(\frac{3}{5}-\frac{2}{5}x\right)$ and then factor it out.
If you combine the above it helps to answer your question. First A, you just have to factor out a $(-1)$ $(3-2x)^4=\bigl((-1)\cdot (-3+2x)\bigr)^4=(-1)^4\cdot (-3+2x)^4=(2x-3)^4$ then use the properties of the power notation and then you have to realise that $(-1)^4=1$ and multiplying with $1$ doesn't change things, because the $1$ is the identity or neutral element with respect to multiplication. Finally you rearrange $-3+2x=2x-3$, this is possible because addition and with it subtraction is commutative $a+b=b+a$.
This works for C too $ \frac{(2x-3)^6}{(3-2x)^2}=\frac{(2x-3)^6}{\bigl((-1)\cdot(-3+2x)\bigr)^2}=\frac{(2x-3)^6}{(-1)^2\cdot(-3+2x)^2}=\frac{(2x-3)^6}{(-3+2x)^2}=\frac{(2x-3)^6}{(2x-3)^2}=(2x-3)^4 $ The last step is just cancellation as in any other fraction. When you do this type of algebraic manipulation more often you will quickly learn these rules by heart. And then realize more general and powerful patterns. For example this swapping $(a-b)^n=(b-a)^n$ we did, works whenever $n$ is even, because we factor out $(-1)^n=1$ and a product with an even number of negative factors is always positive.
As for B it's just an application of the abbreviation rule $a^n=\underbrace{a\cdot a\cdot a\cdot \ldots\cdot a}_{n\text{ times}}$ which can also be written recursively as $a^n=a^{n-1}\cdot a\qquad a^1=a$ Perhaps you can spot the pattern $a^4=a^3\cdot a\qquad$ as $a$ is just a variable, a general placeholder, we have $(2x-3)^3(2x-3)=(2x-3)^4$
These algebaric manipulations take practice, it's like learning a language or swimming or playing the piano, it takes practice to spot the patterns. For someone more practised the derivations above are in painfully slow motion.