The probability that one random TV is defected is $p=1/100$; the probability that it isn't is correspondingly $1-p$.
If we assume that the probability of one TV being defected is independent from the probability of another TV being defected (i.e. you only assume random failures, not e.g. a machine defect which makes a whole bunch of TVs defective at once), the peobabilities for different TVs just multiply. For example if you draw three TVs, the probability that the first TV is OK, the second defected and the third OK again is $(1-p)p(1-p)=9801/1000000$ or approximately 1%.
Now if you ask what is the probability that exactly $k$ TVs out of $n$ are defected, then you have several cases ecluding each other. For example, the first $k$ TVs could be defective, and all following are not. Or only the last $k$ are defective. Or the 1st, 3rd, 5th, …, $(2k-1)$th are defective. In total there are ${n \choose k} = \frac{n!}{k!(n-k)!}$ different possibilities. To get the total probability, you have to add up the probability for each single case. However in each single case, you have $k$ factors $p$ (because $k$ TVs are defective) and $n-k$ factors $1-p$ (because $1-p$ TVs are OK). Therefore your probability is just ${n\choose k}p^k(1-p)^{n-k}$. This is known as binomial distribution, because it's just one term in the binomial formula of $(p+q)^n$ where $q=1-p$ (which BTW immediately shows that the probabilities sum up to 1 as they should, because $(p+(1-p))^n=1^n=1$.
Now to your specific questions. You have a batch of 10 TVs, therefore $n=10$.
(a) You ask for the probability that one TV is defected. This can be interpreted in two ways: (a1) What is the probability that exactly one TV is defected? Or (a2) What is the probability that at least one TV is defected?
For (a1) it is now just a matter of inserting into the formula above. Since ${n \choose 1}=n$, we get $P_{a1}=10 p(1-p)^9 \approx 0.091 = 9.1\%$.
For case (a2), the probability that at least one TV is defected is one minus the probability that none is defected. But the probability for the latter is also easily calculated with the binomial distribution: Since ${n\choose 0}=1$, we get $P_{a2}=1-(1-p)^10\approx 0.096 = 9.6\%$
For case (b), note that at most two TVs are defected if either non, exactly one, or exactly two TVs are defected. Since those possibilities exclude each other (it's not possible that you have both exactly one and exactly two defected TVs in your sample), you just have to add those probabilities. Using ${n\choose 2}=\frac{n(n-1)}{2}$, we therefore get $P_b = (1-p)^10+10p(1-p)^9+45p^2(1-p)^8 \approx 0.9999 = 99.99\%$.