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.
I have used a different method for confirmation, converting a straight line to a $2D$ lattice path.
Number of tosses taken has to be of the form $(2k+1)$, i.e. odd
$k$ can range from $0$ to $\infty$
in $(2k+1)$ tosses, there will be $(k+1)$ "$p\;$steps" and $\;k\;$ "$q$ steps" $(p= 0.7)$
$$\text{Thus the formula} = \sum_{k=0}^\infty \binom{2k+1}{k+1}p^{k+1}q^k$$
Wolfram shows that the sum converges,
and confirms the answer $= 2.5$
PS: Explanation of the Formula
The basic idea is to conceive the problem as one of a $2D$ lattice path rather than a straight line.
If equality is achieved at the $(2k+1)^{\text{th}}$ toss, the first $2k$ tosses must be such that the number of heads is always strictly less than tails till that point, and equality achieved only on the last toss. The number of such arrangements is the well known Catalan Number and is equal to $\dfrac{1}{k+1}\dbinom{2k}{k}$.
Thus with equality being achieved at the $(2k+1)^{th}$ toss, the expected number of tosses is:
$$ = \sum_{k=0}^{\infty}\frac{2k+1}{k+1}\binom{2k}{k}0.7^{k+1}0.3^{k}$$
which further simplifies to the formula used in the body.
Best Answer
I think you're right. Another way to see it is as follows: consider a biased random walk on the integers. The walk starts at 1. The question is what is the expected time until it hits 0. Let $x $ be that expectaion. Then from any position $n>0$ the expected time to hit $0$ is $nx $; the walk must make $n $ steps to the left. Therefore, $x $ must satisfy the equation $$ x=1+0.3*2x .$$ When starting at 1, after 1 step the walk either hits 0 with probability $.7$ or moves to 2 with probability $.3$. In the former event the (expected) hitting time is 1; in the latter event it is $1+2x $.