Step 1: We know $\tan 45^{\circ}=1$. Given $\tan(A+B)=1$, so
$$A+B=45^{\circ} \quad \cdots (i)$$
Step 2: We know $\tan 30^{\circ}=\dfrac{1}{\sqrt{3}}$. Given $\tan(A-B)=\dfrac{1}{\sqrt{3}}$, so
$$A-B=30^{\circ} \quad \cdots (ii)$$
Step 3: Add (i) and (ii).
$$2A=75^{\circ} \;\Rightarrow\; A=37.5^{\circ}$$
Step 4: Subtract (ii) from (i).
$$2B=15^{\circ} \;\Rightarrow\; B=7.5^{\circ}$$
Hence $A=37.5^{\circ}$ (i.e. $37^{\circ}30'$) and $B=7.5^{\circ}$ (i.e. $7^{\circ}30'$).