To #1:
$\displaystyle \dfrac x{x-3}\leq2$ is not defined for x = 3.
$\displaystyle \dfrac x{x-3}\leq2~\implies~\dfrac x{x-3} - \dfrac{2(x-3)}{x-3}\leq 0$
Simplify the LHS to one quotient.
A quotient is smaller than zero - that means negative - if numerator and denominator have different signs. This set up will lead to a chain of inequalities which are easy to solve. Go ahead!
According to my previous post you should have got
$\displaystyle \dfrac{-x+6}{x-3}\leq 0~,~x\ne 3$
First case: numerator positive and denominator negative:
$\displaystyle -x+6\geq 0 ~\wedge~x-3<0$. Here you should come out with x < 3 or in interval notation $\displaystyle (-\infty, 3)$
Second case numerator negative and denominator positive:
$\displaystyle -x+6\leq 0 ~\wedge~x-3>0$. Here you should come out with $\displaystyle x \geq 6$ or in interval notation $\displaystyle [6, +\infty)$
Captain Black's point is that since you are given choices for possible answers, you don't need to solve the inequality. Just put in values from the given intervals to see if they satisfy the inequality or not. For example, if (a) $\displaystyle (-infty, 3)\cup (6, \infty)$ is the correct answer then it should be true that no number between 3 and 6 satisfies the inequality while numbers less than 3 and larger than 6 do satisfy it. Just try some values. Of course, you can't be certain that, just because you a value you pick does satisfy the inequality, all in that interval will, but if you find a value that does NOT, then you know it is not true that all numbers in the interval satisfy it.
If you really would like to solve the inequality, use the fact that inequalities like this can change from "<" to ">" only where they are "=" or are not defined. Now the first one $\displaystyle \frac{x}{x- 3}$ is undefined at x= 3 while if $\displaystyle \frac{x}{x- 3}= 2$, then $\displaystyle x= 2(x- 3)= 2x- 6$. Adding 6 to both sides and subtracting x from both sides, x= 6. The two numbers, 3 and 6, are the only values where "<" can change to ">" and vice versa. Checking one value in x< 3, 3< x< 6, and 6< x will tell you whether all the numbers in each interval give ">" or "<".
For the second, the function is always define and we have "=" only if $\displaystyle -2x^2+ 5x= -12$ which is the same as $\displaystyle 2x^2- 5x- 12= (2x+ 3)(x- 4)= 0$. Again that gives 2 points where the inequality can change so 3 intervals in which it must be a specific way for all numbers in that interval.