Question 1. If $X$ is a random variable that counts the number of heads obtained in $n = 2$ coin flips, then we are given $\Pr[X \ge 1] = 2/3$, or equivalently, $\Pr[X = 0] = 1/3 = (1-p)^2$, where $p$ is the individual probability of observing heads for a single coin flip. Therefore, $p = 1 - 1/\sqrt{3}$.
Next, let $N$ be a random variable that represents the number of coin flips needed to observe the first head; thus $N \sim {\rm Geometric}(p)$, and we need to find the smallest positive integer $k$ such that $\Pr[N \le k] \ge 0.99$. Since $\Pr[N = k] = p(1-p)^{k-1}$, I leave the remainder of the solution to you as an exercise; suffice it to say, you will definitely need more than 3 coin flips.
Question 2. Your answer obviously must be a function of $p$, $n$, and $k$. It is not possible to give a numeric answer. Clearly, $X \sim {\rm Binomial}(n,p)$ represents the number of blue balls in the urn, and $n-X$ the number of green balls.
Next, let $Y$ be the number of blue balls drawn from the urn out of $k$ trials with replacement. Then $Y \mid X \sim {\rm Binomial}(k,X/n)$. You want to determine $$\Pr[X = n \mid Y = k] = \frac{\Pr[Y = k \mid X = n]\Pr[X = n]}{\Pr[Y = k]}.$$ It is trivial to see that $\Pr[Y = k \mid X = n] = 1$, and $\Pr[X = n] = p^n$. The denominator is the tricky part: You need to write $$\Pr[Y = k] = \sum_{x=0}^n \Pr[Y = k \mid X = x]\Pr[X = x],$$ by the law of total probability. Again, I leave the remainder of the solution for you to work out.
Your expression for $P(x = 20)$ looks good to me.
Hint For the other question: The complement of the event "at least one head" is the event "no heads".
Best Answer
Let $P(H) = p$ be the probability of one head. In many scenarios, this probability is assumed to be $p = \frac{1}{2}$ for an unbiased coin. In this instance, $P(H) = 3P(T)$ so that $p = 3(1-p) \implies 4p = 3$ or $p = \frac{3}{4}$.
You are interested in the event that out of three coin tosses, at least 2 of them are Heads, or equivalently, at most one of them is tails. So you are interested in finding the likelihood of zero tails, or one tails.
The probability of zero tails would be the case where you only received heads. Since each coin toss is independent, you can multiply these three tosses together: $P(H)P(H)P(H) = p^3$ or in your case, $(\frac{3}{4})^3 = \frac{27}{64}$.
Now we must consider the case where one of your coin flips is a tails. Since you have three flips, you have three independent opportunities for tails. The likelihood of two heads and one tails is $3(p^2)(1-p)$. The reason for the 3 coefficient is the fact that there are three possible events which include two heads and one tails: $HHT, HTH, THH$. In your case (where the coin is 3 times more likely to have heads): $3(\frac{3}{4})^2(\frac{1}{4}) = \frac{27}{64}$.
Adding those events together you get $p^3 + 3(p^2)(1-p) = \frac{54}{64}$. Note that the 3 coefficient is a result of the binomial expansion as used by @Back2Basic