We'll first prove the hint.
Let $0 .
For this, we use the mean value theorem in $[a,b]$.
Why does $f$ satisfy the hypothesis of MVT on $[a,b]$?
As $f$ is differentiable on $(0,\infty)$, $f$ is continuous on $[a,b]$. And, $f$ is, in fact, differentiable on $[a,b]$, if you define differentability at endpoints using one-sided derivatives.
So, there exists $c \in (a,b)$ \dfrac{f(b)-f(a)}{b-a}=f'(c) >\dfrac{1}{c}>\dfrac{1}{b} Now, noting that, $b-a>0$, we can actually take $b-a$ to the other side, taking us to $f(b)-f(a)>\dfrac{b-a}{b}$
This proves your hint.
Now to prove that, $f$ is not uniformly continuous, we actually prove the contrapositive of the definition:
Contrapositive of uniform continuity
$f$ is NOT uniformly continuous if there exists $\epsilon>0$ such that for each $\delta >0$ there exists $x,y$ such that $|x-y|<\delta$ but $|f(x)-f(y)|\geq \epsilon$.
I think this should not be hard. So, how do we go about this? Set $b=2a$. Note that this is consistent with our assumption on $a$ and $b$ (i.e. $a < b $). There exists an $\epsilon>0$, here $\epsilon =\dfrac{1}{2}$, such that for any $\delta>0$, here any $a>0$, the claimed inequality holds.
So, we are through.