Denote by $D$ the mid-point of $BC$. We offer two ways of doing part (a).

First way:
Continue the line segment $AD$ through $D$ to the point $A'$ chosen so that $|A'D| = |DA| = m_a$. Consider the triangles $A'DC$ and $ADB$. Note that $\angle A'DC = \angle ADB$, $|BD| = |DC|$ and $|A'D| = |AD|$, by construction.

Hence, these triangles are congruent and so, in particular, $|A'C| = |AB|$.
Now consider the triangle $A'CA$, and apply the triangle inequality to infer that
$$2m_a = |A'D| + |DA| = |A'A| < |CA| + |A'C| = b + c.$$

Second way:
Two applications of the Cosine Rule tell us that
$$am_a \cos \angle BDA = m_a^2 + \left(\frac{a}{2}\right)^2 - c^2 \quad \text{and} \\ -am_a \cos \angle BDA = am_a \cos \angle CDA = m_a^2 + \left(\frac{a}{2}\right)^2 - b^2.$$
Thus, eliminating $\cos \angle BDA$, we see that
$$4m_a^2 = 2(b^2 + c^2) - a^2.$$
Hence $2m_a < b + c$ iff
$$2(b^2 + c^2) - a^2 < b^2 + 2bc + c^2 \iff b^2 + c^2 - a^2 < 2bc \iff \cos(\angle BAC) < 1,$$
which is true. In like manner, $2m_b < c + a$, $2m_c < a + b$, and so
$$2m_a + 2m_b + 2m_c < (a+b) + (b+c) + (c+a) = 2(a+b+c),$$
i.e. $m_a + m_b + m_c < a + b + c$. This completes the proof of part (a).

b.
The numbers $2$, $4$, $5$ are the side lengths of a triangle $ABC$ with $a=4$, $b=2$, $c=5$ for which
$$m_a^2 = \frac{2(b^2 + c^2) - a^2}{4} = \frac{21}{2} > 10 = bc.$$
Hence (i).

(ii) occurs in any triangle in which $b = c$, because, in such a case,
$$bc = b^2 = m_a^2 + \left(\frac{a}{2}\right)^2 > m_a^2.$$