$\tan (x + y) = \dfrac{\tan x + \tan y}{1 - \tan x \, \tan y}$
$\tan (x + y) = \dfrac{\frac{1}{3} + \frac{1}{2}}{1 - \frac{1}{3}(\frac{1}{2})}$
$\tan (x + y) = \dfrac{\,\frac{5}{6}\,}{\frac{5}{6}}$
$\tan (x + y) = 1$ answer
$\tan (x - y) = \dfrac{\tan x - \tan y}{1 + \tan x \, \tan y}$
$\tan (x - y) = \dfrac{\frac{1}{3} - \frac{1}{2}}{1 + \frac{1}{3}(\frac{1}{2})}$
$\tan (x - y) = \dfrac{\,-\frac{1}{6}\,}{\frac{7}{6}}$
$\tan (x - y) = -\frac{1}{7}$ answer