Question 1:
I'm asked to prove that there exists an $n\in\mathbb{N}$ such that $$\frac{1}{n+1}\leq\frac{a}{b}<\frac{1}{n},$$ where $0<\frac{a}{b}<1$. Here $\frac{a}{b}$ is a fraction in lowest terms.
Q1(Thoughts):
If $\frac{1}{n+1}\leq\frac{a}{b}<\frac{1}{n},$ then with a bit of algebra we get that $$0\leq\frac{an^2-bn+a}{b}<1,$$ so since $0<\frac{a}{b}<1$ we can set $an^2-bn+a=a$ to get that $an^2-bn=0$, which produces $$n=\frac{b\pm\sqrt{b^2-4a\cdot 0}}{2a}=\frac{b\pm b}{2a},$$ and since $n\neq 0$, then we get that $n=\frac{b}{a}$.
OK, so that was totally wrong, but I've got that $\frac{a}{b}\in[\frac{1}{2},1)\implies n=1,$ $\frac{a}{b}\in[\frac{1}{3},\frac{1}{2})\implies n=2$, and so on ad infinitum, but I don't know that this is a proof. It looks as though no matter what $\frac{a}{b}\in\bigcup^{\infty}_{n=1}[\frac{1}{n+1},\frac{1}{n}]$.
Question 2:
If $n$ is chosen as in Q1, prove that $\frac{a}{b}-\frac{1}{n+1}$ is a fraction that in lowest terms has a numerator less than $a$.
Q2(Thoughts):
I think this might have something to do with $n<\frac{b}{a}$.
I'll be adding more to this as I go, but my question is essentially something like "Is this right?"
There's a problem with your proof, since you conclude that $n = \frac b a$, which is not guaranteed to be an integer. Furthermore, as Michael pointed out, the conclusion that $n = \frac b a$ leads to $\frac a b < \frac a b$, which is problematic.
Regardless, there's actually an easier way to proceed, using the Archimedian property. Let $\alpha = \frac a b$, so that $\frac 1 \alpha > 1$. Let $n$ be the greatest natural number less than $\alpha$, so that $n + 1 \ge \alpha$. What can you conclude from here?