There are errors of the same kind in i) and ii). In addition, the first two parts of iii) were done inefficiently, and there were a couple of numerical issues.
i) You were asked for the probability that at least $80$% in a random sample of $5$ use the machine. This is
$$\binom{5}{4}(0.8)^4(0.2)^1 +\binom{5}{5}(0.8)^5(0.2)^0.$$
ii) The same type of mistake was made. We want the probability of exactly $8$ plus the probability of exactly $9$ plus the probability of exactly $10$.
iii-a) Same answer as for i). You did it the harder may, it is easier to add together probability of $4$, probability of $5$.
iii-b) Same answer as for ii). Again, you did it in much too hard a way.
iii-c) The distribution approaches a normal distribution ("bell-shaped curve") which is symmetric about the mean. So the probability that $X$ is greater than or equal to the mean approaches $1/2$.
Comments on computation: For iii-a), you accidentally wrote down the wrong probability. I get $0.73728$. Two errors here, you meant to subtract from $1$, you did, but then the two numbers were somehow transposed. There is also an error probably due to rounding. There is also a typo in which you wrote down the wrong exponent, but did the calculation right. For iii-b), and therefore for ii), I get $0.6777995$, which differs from your answer in the fourth decimal place. In the real world, no bih deal, the numbers like $80$\% are presumably not exact. But you should use the full capacities of your calculator, in particular the memory feature, to avoid rounding errors that add up. It will also save you time, and maybe cut down on mistakes. Rekeying takes time, and keying errors are easy to make.
The probability being computed is the probability that the attacker will catch up. We can call that event $A$ so we are computing $P(A).$
The way we are computing it is via the law of total probability. In other words, splitting it up into cases. The cases chose are how far along in their attack the attacker happens to be after $z$ blocks have been added. We let $B_k$ be the event that the attacker has made $k$ blocks of progress. We are assuming (more on this later) that $P(B_k) = \frac{1}{k!}\lambda^k e^{-\lambda}.$ We know that given that the attack has made $k$ blocks, the probability they catch up is (1) $1$ if they have already caught up (i.e. $k\ge z)$ (2) $(q/p)^{z-k}$ if they haven't caught up and are $z-k$ blocks behind.
Then we just decompose by the law of total probability: $$P(A) = \sum_k P(A|B_k)P(B_k) $$ where $P(B_k)=\frac{1}{k!}\lambda^k e^{-\lambda}$ and $P(A|B_k) = (q/p)^{z-k}$ for $k<z$ and $P(A|B_k) = 1$ for $k>z.$
So the probability they catch up is the probability they haven't got any blocks yet and they catch up, plus the probability that they've made one block and they catch up, plus the probability they've made two blocks and they catch up, etc. Note that as $k$ increases $P(A|B_k)$ (the probability the catch up given that they have made $k$ blocks) increases, eventually becoming $1,$ while $P(B_k)$ (the probability they've made $k$ blocks when the honest nodes have made $z$) decreases.
Now for the part about binomials. We can consider both the honest network and the attacker to be independent poisson processes, producing blocks at rate $\lambda_p$ and $\lambda_q$ so that the probability the next block is produced by honest network is $p=\frac{\lambda_p}{\lambda_p+\lambda_q}$ and similarly $q=1-p$ for the attacker. In this case the distribution of number of blocks $k$ the attacker has produced is negative binomial. This is because it is the number of successes for the attacker before $z$ failures (where a failure is the honest network producing a block). This seems to me a reasonable model of what's going on.
So we should instead have $$ P(B_k) = {k+z-1\choose k } q^k(1-q)^z.$$
Now, it happens sometimes that a negative binomial (or a binomial) can be approximated by a Poisson distribution. When is this true? Well, in general, it's when you have a lot of a attempts with a small probability. This would be the case when $z$ is large and $q$ is very small. This doesn't seem like a particularly natural limit for this problem as we'd imagine moderate sized $z$ and $q$ being of interest. An answerer Bob at the other site says something to the contrary but they're confusing the trials of the nonce for the trials to get the next block. The former process is actually a (not negative) binomial distribution which is a poisson process to a good approximation.
What Satoshi does is assume that the honest blocks take exactly their average time $$T_z = z/\lambda_p$$ to get the $z$ blocks. Then since we're still assuming the attacker generates blocks in a Poisson process, the number of blocks the attacker creates in that time is Poisson distributed with mean $$\lambda_q T_z = \frac{\lambda_qz}{\lambda_p} = \frac{q}{p}z.$$
But it seems a questionable approximation to suddenly have the honest nodes be deterministic while the attacker still has fluctuations. Of course, there will be a limit where this is effectively true, and it's in exactly the same limit as where the negative binomial above can be approximated as a Poisson.
Best Answer
Hint. Try to use this property $$P(\text{x to happen})=1-P(\text{x not to happen})$$
Thus, if the probability of catching a fish in one try is $p$ then
Then the probability to catch at least one fish in $n$ tries is $$1-(1-p)^n$$ Now, the question is to find minimal $n$ such that $$1-(1-0.8)^n\geq 0.9$$