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.
Best Answer
The answer you got is correct and so is the method.
If you want another method, you can take all the cases $^{15}C_6$ and subtract from them the cases where $0$ girls are chosen $^5C_0 \cdot$ $^{10}C_6$ and where $1$ girl is chosen $^5C_1 \cdot ^{10}C_5$ which also gives $3535$ as the answer.