How large must x be so that something happens with probability at least n? I'm having trouble understanding the answers to the below questions. I have provided my own attempts at the solutions below the

i) I set up the problem like $0.5^x \geq 0.9$ because it reads, "the probability of heads appearing $x$ times is at least $0.9$. I'm not sure why the book wrote the problem the other way around?
ii) Like the i) the book takes the complement for some reason
iii) Same confusion as i) and ii)
Thanks for any clarification.
EDIT: 
The answers below have started helping me understand this problem. However, I have a question on how the math actually works out.
i. $(1 - 0.5)^x \geq 0.9$ is that how to set the problem up? 
 A: The reason for taking the compliment is as follows. Let's look at (i) for example.
Flip a coin. The probability of getting heads is 1/2, and the probability of tails is 1/2.
Now flip it again. The probability of:


*

*Getting two heads is (1/2)(1/2)= 1/4

*Two tails is (1/2)(1/2)= 1/4)

*Heads then tails is (1/2)(1/2) = 1/4

*Tails then heads is (1/2)(1/2) = 1/4


Of these possibilities, three of them contain a head with a total probability of 3/4, which is still less than 0.9. So we would have to flip another coin.
In other words, after x flips, the only way NOT to get at least 1 heads is to only roll tails. Since the probability of rolling tails is 1/2, you want to find the number of consecutive rolls of tails that has a probability of less than 0.1.
A minor point is to round x to 4 and 13 since you can only flip a coin or roll a die an integer number of times.
A: They set up the complement case because "at least one" is the complement of "none". Otherwise you would have to sum up all the cases: $1$ heads, $2$ heads, ... $x$ heads.
In other words, you start with
$$\mathbb P(\text{at least one head}) \geq 0.9 $$
and just substitute $\mathbb P(\text{at least one head}) = 1 - \mathbb P(\text{no head}) $.
Also note that $0.5^x \geq 0.9$ doesn't have any solutions for $x \geq 1$ so that's clearly not the correct approach.  
