Correct me if I am missing something or show me the better way.
Let $0<a<b$ and consider the sequence of functions $$f_{n}(x)=\frac{1-(x/b)^{n}}{1+(a/x)^{n}}$$ for $n\in \mathbb{N}$.
i) First consider $x\in [a,b]$. Show that $f_{n}(x)$ does not converge uniformly on this set.
ii) Now reduce the domain into $x\in [a+\epsilon, b-\epsilon]$ where $\epsilon>0$, and it is assumed that $\epsilon<(b-a)/2$. Prove that $f_{n}(x)$ is uniformly convergent.
My answer to i) I have shown that $f_{n}(x)$ converges pointwise to
$$ f(x)=\begin{cases} \frac{1}{2} & \text{ if } a=x \\ 1 & \text{ if } a<x<b \\ 0 & \text{ if } x=b. \end{cases} $$ Since, for example, $\lim_{x\to a^{-}}f(x)=1\neq f(a)$, so $f$ is not continous on $[a,b]$. Thus $f$ does not converge uniformly on $[a,b]$. Is it enough to answer like that?
My answer to ii) I believe it says proving that $$ \lim_{n\to\infty}\sup_{x\in [a+\epsilon, b-\epsilon]}\left | f(x)-f_{n}(x) \right |=0. $$ First, we know that $1-\left ( \frac{x}{b} \right )^{n}\geq 1-\left ( \frac{b-\epsilon}{b} \right )^{n}$ and $1+\left ( \frac{a}{x} \right )^{n}\leq 1+\left ( \frac{a}{a+\epsilon} \right )^{n}$ for all $x\in [a+\epsilon, b-\epsilon]$. Since $[a+\epsilon, b-\epsilon]\subseteq (a,b)$, so
\begin{align*}\left | f(x)-f_{n}(x) \right |&\leq \left | 1-\frac{1-\left ( \frac{b-\epsilon}{b} \right )^{n}}{1+\left ( \frac{a}{a+\epsilon} \right )^{n}} \right |=\left | \frac{\left ( \frac{a}{a+\epsilon} \right )^{n}-\left ( \frac{b-\epsilon}{b} \right )^{n}}{1+\left ( \frac{a}{a+\epsilon} \right )^{n}} \right |\\ &\leq \left | \left ( \frac{a}{a+\epsilon} \right )^{n}-\left ( \frac{b-\epsilon}{b} \right )^{n} \right |\leq\left ( \frac{a}{a+\epsilon} \right )^{n}+\left ( \frac{b-\epsilon}{b} \right )^{n}.\end{align*} Since $a+\epsilon>a>0$ and $0<b-\epsilon<b$, we finally get $\left | f(x)-f_{n}(x) \right |\to 0$ as $n\to\infty$.
I do not know if I should use something like that for all $\epsilon>0$ there exist a $N\in\mathbb{N}$ such that $\sup_{x\in [a-\epsilon,b+\epsilon]}\left | f(x)-f_{n}(x) \right |<\epsilon$ for all $n\geq N$.
Your answer to (i) is fine.
Your proof for (ii), in order to make use of the $\varepsilon$-property which in my case is a $\delta$-property as $\varepsilon$ already has been used, could be extended as follows:
Note that for $f_n$ to converge uniformly on $[a+\varepsilon, b-\varepsilon ]$ to $f$ you have to show, that for each $\delta > 0 $ there exists an $n_0 \in \mathbb{N}$ such that for every $n \ge n_0$ and every $x \in [a+\varepsilon, b-\varepsilon]$ it holds that $\lvert f(x) - f_n(x) \rvert \lt \delta$.
Proof:
Let $\delta > 0$. For every $\varepsilon > 0$ the sequences defined as
$$a_n := \left( \frac{a}{a + \varepsilon}\right)^n \quad \text{ and } \quad b_n := \left( \frac{b - \varepsilon}{b}\right)^n $$ are null sequences, i.e. there exists $n_a \in \mathbb{N}$ such that for every $n \ge n_a$
$$ \lvert a_n \rvert < \frac{\delta}{2}.$$
Analogously there exists $n_b \in \mathbb{N}$ such that for every $n \ge n_b$
$$ \lvert b_n \rvert < \frac{\delta}{2}.$$
Now let $n_0 := \max(n_a,n_b)$. Performing the estimates in your proof for all $n \ge n_0$ and all $x \in [a+\varepsilon, b-\varepsilon ]$ the following holds
$$ \lvert f(x) - f_n(x) \rvert \leq \left( \frac{a}{a + \varepsilon}\right)^n + \left( \frac{b - \varepsilon}{b}\right)^n \lt \frac{\delta}{2} + \frac{\delta}{2} = \delta. $$