Evaluate the expression under the given conditions.

tan(θ+φ)

given cosθ = -1/3 in Q3
and sinφ = 1/4 in Q2

I'm not really sure where to begin.

If $\cos{\theta} = -\frac{1}{3}$ in quad III, what is $\sin{\theta}$?

If $\sin{\phi} = \frac{1}{4}$ in quad II, what is $\cos{\phi}$?

get those two values and you can determine the tangent of each angle since $\tan{\theta} = \dfrac{\sin{\theta}}{\cos{\theta}}$ and $\tan{\phi} = \dfrac{\sin{\phi}}{\cos{\phi}}$

finally, you can use the sum identity for tangent ...

$\tan(\theta + \phi) = \dfrac{\tan{\theta} + \tan{\phi}}{1 - \tan{\theta}\tan{\phi}}$

$tan(\theta+\phi)=\dfrac{tan\theta+tan\phi}{1-tan\theta tan\phi}\;\;\;\;\;\left(\because tan(A+B)=\dfrac{tanA+tanB}{1-tanA\,tanB}\right)$
$=\dfrac{\dfrac{sin\theta}{cos\theta}+\dfrac {sin\phi}{cos\phi}}{1-\dfrac{sin\theta}{cos\theta}\dfrac{sin\phi}{cos \phi}}$
$=\dfrac{sin\theta cos\phi+cos\theta sin\phi}{cos\theta cos\phi-sin\theta sin\phi}$
$\because cos\theta=\dfrac{-1}{3}\;\;$ $\therefore sin\theta=\sqrt{1-(cos)^2 \theta}=\sqrt{1-\frac{1}{9}}=\sqrt{\frac{8}{9}}=\dfrac{2\sqrt2}{3}$
similarly $\because sin\phi =\dfrac{1}{4}\;\; \therefore cos\phi=\dfrac{\sqrt{15}}{4}$
$tan(\theta + \phi)=\dfrac{\dfrac{2\sqrt2}{3} \dfrac{\sqrt{15}}{4}+\dfrac{-1}{3} \dfrac{1}{4}}{\dfrac{-1}{3}\dfrac{\sqrt{15}}{4}-\dfrac{2\sqrt2}{3}\dfrac{1}{4}}$
$=\dfrac{2\sqrt{30}-1}{-\sqrt{15}-2\sqrt{2}}$