I draw bulbs one at a time until I get a $75$ watt bulb. Because I am easily irritated, I will be unhappy if I have to draw $2$ or more lightbulbs. The question asks for the probability I will be unhappy.
The probability I will be unhappy is $1$ minus the probability I will be happy. And I will be happy only if I get a $75$ watt bulb immediately. The probability of this is $\frac{4}{15}$, so the probability I will be unhappy is $\frac{11}{15}$.
In terms of events, the number of bulbs I draw until we I a $75$ watt bulb is any of $1,2,3,4,\dots, 12$ (I assume I am drawing without replacement). The event $A$ that I will be unhappy is the set $\{2,3,4,\dots,12\}$. The event I will be happy is the complement $A'$ of $A$, it is the set $\{1\}$.
Or else we can think of the sample space as consisting of the following "words:" S, FS, FFS, FFFS, FFFFS, and so on where for example FFFS means I got a weak bulb three times in a row, and then got a $75$ watt bulb on the fourth try. Here F means failure and S means success. The event $A$ consists of FS, FFS, FFFS, and so on, everybody but just plain S.
It is fairly often the case that to find the probability $\Pr(A)$ of an event $A$, it is easier to first find $\Pr(A')$, and then use the fact that $\Pr(A)=1-\Pr(A')$.
Remark: If the explanation given is the one in your second paragraph, then the explanation is not good. The event $A$ is the event that at least two (two or more) bulbs are selected.
The experiment consists of selecting bulbs until we get a $75$ watt bulb, and then stopping. So selecting $0$ bulbs is not one of the possible outcomes of the experiment.
As to the "at most" part, you have not described an explicit problem. But in the context of the lightbulb problem, "at most $3$" means $1$ or $2$ or $3$.
Best Answer
I cannot find a flaw in your solution of (a).
Let $X$ denote the number of $75$-W bulbs that are selected.
Then:$$P(X=2\mid X\geq1)=\frac{P(X=2)}{P(X\geq1)}=\frac{P(X=2)}{1-P(X=0)}$$
Here $P(X=2)=\frac6{15}\frac5{14}$ and $P(X=0)=\frac9{15}\frac8{14}$ which agrees with what you wrote as solution.
Also your answer on (b) is correct.
If $Y$ denotes the number of $40$-W bulbs and $Z$ the number of $60$-W bulbs that are selected and $E$ denotes the event that both selected bulbs have the same rating then:$$P(E\mid X\leq1)=\frac{P(E\cap\{X\leq1)\}}{P(X\leq1)}=\frac{P(E\cap\{X=0\})+P(E\cap\{X=1\})}{1-P(X=2)}=$$$$\frac{P(E\cap\{X=0\})+0}{1-P(X=2)}=\frac{P(Y=2)+P(Z=2)}{1-P(X=2)}$$