Edit: Thanks to Jonas for pointing out a mistake, and for suggesting an improvement.
(1) Let $\epsilon > 0$. Choose $\delta = \epsilon/K$. Then if $|x-y|<\delta$, then $|f(x)-f(y)|. Thus $f$ is uniformly continuous.
(2) $f(x)=\sqrt{x}$ on $[0,1]$. To see that $f$ is not an $\mathcal{L}$-function, note that since $f$ is differentiable with derivative $f'(x)=\frac{1}{2\sqrt{x}}$, the difference quotient $\displaystyle \frac{|f(x)-f(y)|}{|x-y|}$ is unbounded. But $f$ is uniformly continuous on $[0,1]$:
For $x=0$ or $y=0$ the problem is trivial. For $x,y>0$, $|\sqrt{x}-\sqrt{y}|\leq \sqrt{|x-y|}$.
Proof: Without loss of generality, take $x > y$. Then $\displaystyle x \leq x+2\sqrt{(x-y)y}=(x-y)+2\sqrt{(x-y)y}+y = (\sqrt{x-y}+\sqrt{y})^2$. Taking the square root of both sides and rearranging, we get the desired inequality.
Now taking $|x-y|<\epsilon^2$, we get $|\sqrt{x}-\sqrt{y}|<\epsilon$.
(3) For any $x,y\in(a,b)$, if $f$ is a $\mathcal{L}$-function then $\displaystyle \frac{|f(x)-f(y)|}{|x-y|}\leq K$, so taking the limit as $y \to x$ we get $f'(x) \leq K$.
(4) False.
If $f(x)=x^{-1}$ on $(0,1)$, then $f'(x)=-x^{-2}$, so $f$ is differentiable on $(0,1)$. But $|f'|$ is obviously unbounded, so by (3) it cannot be a $\mathcal{L}$-function. Thus we have a differentiable function on $(0,1)$ that is not a $\mathcal{L}$-function on $(0,1)$.
If $f(x)=|x|$ on $(-1,1)$, then $f$ is not differentiable at $0$. But $f$ is a $\mathcal{L}$-function with $K=1$, since by the reverse triangle inequality $|~|x|-|y|~| \leq |x-y|$.