The question asks you to find the probability of getting $2$ defective in the sample of $25$ if the true proportion of output that is defective is $1$%, that is, $0.01$. So the correct answer is
$$\binom{25}{2}(0.01)^2(0.99)^{23}.$$
This seems to be roughly $0.0238$.
Part ii) has to do with hypothesis testing. Roughly speaking, we have the Null Hypothesis that indeed the supplier's output is only $1$% defective. We have performed an experiment. Even if the supplier's output is only $1$% defective, by bad luck there could be $2$ defectives in the batch of $25$ that was tested.
But this would not happen very often. If the Null Hypothesis holds, then only about $2.38$% of batches of $25$ would have exactly $2$ defectives. So on the "$1$%" hypothesis, something rather unusual has happened. Not impossible, but unusual.
In hypothesis testing, it is common to set a significance level, moderately often $5$%, or $1$%. If in testing something happens which (under the Null Hypothesis) would happen with probability less than $0.05$, then at significance level $0.05$, one rejects the Null Hypothesis. So at significance level $0.05$, the experimental result described is enough to reject the Null Hypothesis.
By the way, a real statistician would calculate the probability of getting $\ge 2$ defective under the Null Hypothesis. This is about $0.02576$. Still well under $0.05$. I am somewhat troubled that the problem setter focused on exactly $2$, this is the wrong approach to hypothesis testing.
Further comments: To me, using level of significance $0.05$ seems unsuitable, I would want stronger evidence to reject the Null Hypothesis. After all, if you have a long-established relationship with a supplier, it seems foolish to dump it on weakish evidence. Also, it is hard to select $25$ items really randomly. Even if overall output is only $1$% defective, there can be non-random day to day variation.
More sensible is to consider the test results as a warning that something might be wrong. So I would at least recommend that more testing be done, since the Null Hypothesis certainly did not get a clean bill of health.
Your calculation for part (A) is correct.
Your work for part (B) is essentially correct, but there are two issues: first, your notation is nonstandard and a bit incomplete: one should more properly write, for example, $$\Pr[X = x] = \binom{100}{x} p^x (1-p)^{100-x}, \quad x \in \{0, 1, 2, \ldots, 100\}.$$ Second, since you are given $p = 0.01$, it is best that you substitute the value in, giving $$\Pr[X = x] = \binom{100}{x} (0.01)^x (0.99)^{100-x}, \quad x \in \{0, 1, 2, \ldots, 100\}.$$ The same applies when we write $$X \sim \operatorname{Binomial}(n = 100, p = 0.01).$$ If you are not familiar with the binomial coefficient, there are different notations for it: $$\binom{n}{k} = {}_n C_k = C^n_k = C(n,k) = C_{n,k}$$ are all equivalent ways of writing it.
Your calculation for part (C) is incorrect. You are not asked for an expected value. You want the probability that, for a fixed $y \in \{1, 2, \ldots, 100\}$, your testing stops at bulb $y$. So your answer will be a function $f_Y(y)$ of $y$, and it is a mass function satisfying $$\sum_{y=1}^{100} f_Y(y) = 1.$$
To this end, let's work out an example: What is the probability that you stop testing immediately after the first bulb is tested? Remember, you stop once you find a good bulb, or you've tested all the bulbs. So, stopping after testing the first bulb means you tested it and it was good. If $Y$ is a random variable that represents the number of bulbs tested in this sequential manner, then clearly $$\Pr[Y = 1] = 1 - p = 0.99.$$
Now, what is the probability that you stop immediately after the second bulb is tested? This amounts to testing the first bulb and finding it is defective (or else you would have stopped!), then testing the second and finding it is good. So this is $$\Pr[Y = 2] = p(1-p) = (0.01)(0.99).$$ Extending this reasoning further, we can see that $$\Pr[Y = y] = p^{y-1}(1-p) = (0.01)^{y-1}(0.99), \quad y \in \{1, 2, \ldots, 99\}.$$ This is because in order to stop testing right after the $y^{\rm th}$ bulb, you'd have to see $y-1$ defective bulbs, and then the final $y^{\rm th}$ bulb must be good. But why is this formula true only for $y$ up to $99$? Why not $100$? Because $$\Pr[Y = 100] = p^{100} = (0.01)^{100}.$$ Once $100$ bulbs are tested, there isn't a $101^{\rm st}$ bulb to test--you've exhausted your supply, and although you've not found a good bulb, there are no more to test.
This distribution for $Y$ is what we might call an "upper-modified geometric distribution," something akin to zero-modified distributions.
Best Answer
For formatting use dollar signs (double ones for equations that appear on their own line). One good trick is to just click edit on other peoples' posts and you will see their markup. The way you do chooses is "{10 \choose 2}" and you get (edit my post to see markup) $$ {10\choose 2}$$
(a) You're choosing 2 from the 10 defective ones and 4 from the 20 non-defectives (for a total of six) so it's ${10 \choose 2}{20 \choose 4}.$
(b) Right idea, but aside from the propagated mistake from (a) you forgot a term for 0 defective widgets.
(c) There's a way to do this one intuitively, but it's probably best to be careful. Let $D$ be the number of defective widgets. You want the conditional probability $P(D=3|D\ge 1)$ and a quick computation gives $$P(D=3|D\ge 1) = \frac{P(D = 3,D\ge 1)}{P(D\ge 1)} = \frac{P(D=3)}{P(D\ge 1)} =\frac{P(D=3)}{1-P(D=0)}$$ where the first step was just definition, the second used the fact that if $D=3,$ $D$ is automatically $\ge 1$ and the third used that the only way for the number of defective devices to not be $\ge 1$ is for there to be no defective devices. You can compute the two required probabilities the same way as in the first two parts.