First, we need the probability that among the first five balls, excactly three were black (and therefore two were white). This is a standard hypergeometric probability exercise, and the answer is
$$
\frac{\binom53\binom72}{\binom{12}5} = \frac{35}{132}
$$
Then, the probability that after this, the sixth ball is white. At this point there are seven balls left, and five of them are white. So the probability of drawing one of them as the seventh ball is
$$
\frac{5}{7}
$$
Now just multiply these two probabilities together, and you get your $\frac{25}{132}$.
All $3$ approaches are okay.
Number the red balls by $1,2,3$ and number the black balls by $4,5,6,7$
Then you could say that approach1 is linked with a probability space where: $$\Omega=\{\{i,j\}\mid i,j\in\{1,2,3,4,5,6,7\}\text{ and }i\neq j\}$$
Here $|\Omega|=\binom72=21$ agreeing with the denominator in approach1.
Also you could say that approach3 is linked with a probability space where: $$\Omega=\{\langle i,j\rangle\}\mid i,j\in\{1,2,3,4,5,6,7\}\text{ and }i\neq j\}$$
Here $|\Omega|=7\times6=42$ agreeing with the denominator in approach3.
Concerning approach2 (my favorite) both spaces can be used but actually we do not meet any impact of a choice for a probability space.
If $E_i$ denotes the event that the $i$-th ball is red for $i=1,2$ then we apply:$$P(\text{both balls are red})=P(E_1\cap E_2)=P(E_1)P(E_2\mid E_1)=\frac37\frac26$$
Calculating such probability we are not even aware of any underlying probability space.
Best Answer
Note you're sampling with replacement, so each ball drawn can be treated as an independent trial with $(4/16)$ chance of drawing a white ball.
If two are white, any two of the four can be white, so there are ${4\choose 2 } = 6$ possibilities, each of which have probability $(4/16)^2(12/16)^2$ so the probability that two are white is $6*(4/16)^2*(12/16)^2.$
Similarly if three are white, there are $4$ ways and the probability is $$4*(4/16)^3*(12/16)^1.$$
And the probability all four are white is $(4/16)^4.$
The total probability that two or more are white is the sum $$\frac{6*4^2*12^2 + 4*4^3*12 + 4^4}{16^4} = \frac{17152}{16^4} = \frac{67}{256} $$
Your approach of using $12 \choose 2$, etc is better suited to sampling without replacement.