(i) First, $a\cos{C} + \dfrac{c}{2} = b \implies \sin{A}\cos{C} + \dfrac{1}{2}\sin{C}=\sin{B}$. Because $A+B+C=\pi\implies \sin{B} = \sin{(A+C)} = \sin{A}\cos{C} + \cos{A}\sin{C}$, we have $$\dfrac{1}{2}\sin{C}=\cos{A}\sin{C} \implies \cos{A}=\frac{1}{2} \implies A = \boxed{\dfrac{\pi}{3}}$$ (ii) $b=\dfrac{a\sin{B}}{\sin{A}} = \dfrac{2}{\sqrt{3}}$, $c = \dfrac{2}{\sqrt{3}}\sin{C}$. $a+b+c=1+\dfrac{2}{\sqrt{3}}(\sin{B}+\sin{C}) = 1 + \dfrac{2}{3}(\sin{B}+\sin(A+B)) = 1+2\sin(B+\frac{\pi}{6})$ $A=\frac{pi}{3} \implies B\in(0, \dfrac{2\pi}{3}) \implies \sin(B+\dfrac{\pi}{6})\in(\dfrac{1}{2}, 1] \implies \boxed{l \in (2, 3]}$