Trigonometric problem (problem from a Swedish 12th grade ‘Student Exam’ from 1932)

Take $\tan$ of both sides of first equation, $$\tan (\alpha + \beta) = \dfrac{\tan \alpha + \tan \beta}{1-\tan \alpha \tan \beta}$$

$$\Rightarrow \tan 135 = -1 = \dfrac{5}{1-\tan \alpha \tan \beta} $$

$$\Rightarrow \tan \alpha \tan \beta=6$$

Setting $\tan \alpha = x$, $\tan \beta = y$ we get two equations : $$x+y=5 \quad xy=6$$

I believe you can easily solve from here.

Recall that $\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$. Now, given $\alpha+\beta=135^{\circ}$ and $\tan(\alpha+\beta)=\tan 135^\circ=-1$, you get $\frac{5}{1-\tan\alpha\tan\beta}=-1$, i.e. $\tan\alpha\tan\beta=6$. Now, knowing the sum ($5$) and the product ($6$) of $\tan\alpha$ and $\tan\beta$, using Vieta formulas we conclude that $\tan\alpha$ and $\tan\beta$ are the solutions of the quadratic equation $x^2-5x+6=0$, i.e. $\tan\alpha$ and $\tan\beta$ are $2$ and $3$ in some order. This is now very easy to finish off: $\alpha=\arctan 2+n\cdot 180^{\circ}$ and $\beta=135^{\circ}-\alpha=\arctan 3-n\cdot 180^{\circ}$, or vice versa.