(α) To each fraction of $A$ of the form $\frac{2\nu - 1}{2\nu + 2}$, $\nu = 1, 2, \dots, 299$, corresponds a fraction from $B$ of the form $\frac{2\nu}{2\nu + 3}$, $\nu = 1, 2, \dots, 299$. Since
$$0 < \frac{2\nu - 1}{2\nu + 2} < \frac{2\nu}{2\nu + 3} \quad \text{for every } \nu \in \mathbb{N}^{*},$$
and so for $\nu = 1, 2, \dots, 299$, by multiplying by parts the above 299 inequalities we obtain $A < B$.

(β) Since $A > 0$, from $A < B$ we get:
$$A^2 < A \cdot B = \frac{1 \cdot 2 \cdot 3}{599 \cdot 600 \cdot 601} < \frac{1}{100 \cdot 599^2} = \frac{1}{5990^2} \Rightarrow A < \frac{1}{5990}.$$