For the following two (02) items: Let (2x + 3y) = (1)/(2) and (3x +...

Let $2x + 3y = u$ and $3x + 2y = v$.

Given $\cos u = \dfrac{1}{2}$ and $-\pi < u < \pi$, the possible values of $u$ are $\dfrac{\pi}{3}$ and $-\dfrac{\pi}{3}$.

Given $\cos v = \dfrac{\sqrt{3}}{2}$ and $-\pi < v < \pi$, the possible values of $v$ are $\dfrac{\pi}{6}$ and $-\dfrac{\pi}{6}$.

Adding the two equations, we get $u + v = 5(x + y) \Rightarrow x + y = \dfrac{u + v}{5}$.

The possible values of $u + v$ are:

$\dfrac{\pi}{3} + \dfrac{\pi}{6} = \dfrac{\pi}{2}$

$\dfrac{\pi}{3} - \dfrac{\pi}{6} = \dfrac{\pi}{6}$

$-\dfrac{\pi}{3} + \dfrac{\pi}{6} = -\dfrac{\pi}{6}$

$-\dfrac{\pi}{3} - \dfrac{\pi}{6} = -\dfrac{\pi}{2}$

Since all four values of $u + v$ are distinct, there are exactly four distinct values for $x + y$.