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 answer is correct. A shorter way of writing it:
Let $H_n$ denote the event that we get $H$ at the $n$-th coin toss. Hence, we want the probability $$P(H_3 \vert (H_1 \cap H_2))=\frac{P(H_1 \cap H_2 \cap H_3)}{P(H_1 \cap H_2)}=\frac{P(H_1 \cap H_2\cap H_3 \vert fair)\cdot P(fair)+ P(H_1 \cap H_2\cap H_3 \vert unfair)\cdot P(unfair)}{P(H_1 \cap H_2 \vert fair)\cdot P(fair)+ P(H_1 \cap H_2 \vert unfair)\cdot P(unfair)}=\frac{(\frac{1}{2})^3 \cdot \frac{2}{3}+1^3\cdot \frac{1}{3}}{(\frac{1}{2})^2 \cdot \frac{2}{3}+1^2\cdot \frac{1}{3}}=\frac{5}{6}.$$
Best Answer
You have the right pieces, but you’ve not put them together correctly. Suppose that you pick the biassed coin: the probability of getting two heads is $\left(\frac23\right)^2$, and the probability of getting two tails is $\left(\frac13\right)^2$, but this means that the probability of getting the same result twice is $\left(\frac23\right)^2+\left(\frac13\right)^2$, not $\left(\frac23\right)^2\cdot\left(\frac23\right)^2$. Either of the outcomes two heads and two tails is a success (meaning both the same), so the probability of a success is the sum of the individual probabilities. You would multiply if you needed both of these things to happen simultaneously – for instance, if you were flipping the coin four times and needed the first two flips to be heads and the last two to be tails.
Similarly, the probability of getting two heads with the fair coin is $\left(\frac12\right)^2$, and so is the probability of getting two tails, so the probability of getting the same result twice is $\left(\frac12\right)^2+\left(\frac12\right)^2$, not $\left(\frac12\right)^4$. Since the probability of picking each coin is $\frac12$, the correct final result is
$$\frac12\left(\left(\frac23\right)^2+\left(\frac13\right)^2\right)+\frac12\left(\left(\frac12\right)^2+\left(\frac12\right)^2\right)\;,$$
or
$$\frac12\cdot\frac59+\frac12\cdot\frac12=\frac5{18}+\frac14=\frac{19}{36}\;.$$