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 $3$.
$\newcommand{\Var}{\operatorname{Var}}$
I call the life time of a particular bulb $X_i$
1) Is fine.
2) There are 100 bulbs in the box, and hence the average is
$$\bar X = \frac{X_1+\dotsb+X_{100}}{100}.$$
Then
$$E[\bar X] =\frac{1}{100}E[X_1+\dotsb+X_{100}] = \frac{1}{100}\cdot100E[X_1] = 2000$$
and
\begin{align*}
\Var(\bar X) &= \Var\left(\frac{X_1+\dotsb+X_{100}
}{100}\right) \\
&= \frac{1}{100^2}\left[\Var(X_1)+\dotsb+\Var(X_{100})\right]\\
&= \frac{2000^2}{100}\\
&= 40000.
\end{align*}
Then the problem becomes
\begin{align*}
P\left(\frac{X_1+\dotsb+X_{100}}{100}>2500\right)&=1-P\left(\frac{X_1+\dotsb+X_{100}}{100}\leq 2500\right)\\
&\approx1-P\left(Z\leq\frac{2500-2000}{\sqrt{40000}}\right)\tag 1\\\
&=1-\Phi(2.5)\tag 2\\
&=0.006209665
\end{align*}
where in $(1)$ $Z$ is a standard normal and we are using a normal approximation, and in $(2)$, $\Phi$ is the standard normal cdf.
3) Let $S = X_1+\dotsb+X_{100}$. Then
$$E[S] = E[X_1+\dotsb+X_{100}] =100\cdot 2000 = 200000$$
and
$$\Var(S) = \Var(X_1+\dotsb+X_{100}) = 100\cdot 2000^2 =20000^2.$$
Then the problem becomes
\begin{align*}
P(S>220000)&=1-P(S\leq 220000)\\
&\approx 1-P\left(Z\leq \frac{220000-200000}{\sqrt{20000^2}}\right)\\
&=1-\Phi(1)\\
&=0.1586553
\end{align*}
Best Answer
Imagine if you will that each and every bulb has a unique id label.
I.e. you have 40watt bulbs: $A_1, A_2, A_3,A_4$, you have 65watt bulbs $B_1,\dots, B_5$, you have 75watt bulbs $C_1,\dots, C_6$.
Notice that every pair of two bulbs is equally likely to be drawn. For convenience, let us assume order doesn't matter. (The same arguments will apply for if we had let order matter and it won't affect the answer either way). We then set as our sample space $S = \{\text{all ways of drawing two bulbs}\}$.
As we are interested in a conditional probability, we restrict our attention to a subset of the sample space, what I will call $F$, where $F = \{\text{all ways of drawing two bulbs where at least one of them is 75watt}\}$
Let us count how large $F$ is.
There are two cases to consider: either exactly one bulb is 75watt, or both bulbs are 75watt. There are $\binom{6}{1}\binom{5+4}{1}$ number of outcomes for the first case, and there are $\binom{6}{2}$ number of outcomes for the second case. Thus, $|B| = \binom{6}{1}\binom{9}{1} + \binom{6}{2}= 6\cdot 9 + 15 = 69$.
Let $E = \{\text{all ways of drawing two 75watt bulbs}\}$. Notice, we calculated $E$ above, as it was exactly the second case to consider in calculating $F$.
We have then $Pr(E|F) := \frac{|E\cap F|}{|F|} = \frac{|E|}{|F|} = \frac{15}{69}=\frac{5}{23}$
note: this version of the definition of probability only works in an equiprobable setting like this one
If you were to have let order matter, it might be clearer why this gives the same answer as you calculated with the tree diagram.
When order matters, you have that $|F|$ splits into three cases:
first is 75watt, second is 75watt ($6\cdot 5$)
first is 75watt, second is not ($6\cdot 9$)
first is not, second is 75watt ($9\cdot 6$)
and that $|E|$ corresponds to the first case above with a total of $6\cdot 5$ outcomes.
You have then that $Pr(E|F) = \frac{6\cdot 5}{6\cdot 5 + 6\cdot 9 + 9\cdot 6} = \frac{30}{138} = \frac{5}{23}$