1. Let \(x=\tan A\) and \(y=\tan B\).\
2. Using the addition formula: \[\tan(A+B)=\frac{x+y}{1-xy}=3\] ⇒ \(x+y=3(1-xy)=3-3xy\) …(i)\
3. Using the subtraction formula: \[\tan(A-B)=\frac{x-y}{1+xy}=\frac13\] ⇒ \(3(x-y)=1+xy\) ⇒ \(3x-3y-xy=1\) …(ii)\
4. Rewrite (i) as \(x+y+3xy=3\).\
5. Multiply (i) by 3: \(3x+3y+9xy=9\).\
6. Add this to (ii): \( (3x+3y+9xy)+(3x-3y-xy)=9+1\) ⇒ \(6x+8xy=10\) ⇒ \(x(3+4y)=5\) ⇒ \(x=\frac{5}{3+4y}\) …(iii)\
7. Substitute (iii) into (i): \[\frac{5}{3+4y}+y+3y\frac{5}{3+4y}=3\]\
8. Combine the fractions: \[\frac{5+15y}{3+4y}+y=3\] ⇒ \[\frac{5(1+3y)}{3+4y}+y=3\]\
9. Multiply by \(3+4y\): \[5(1+3y)+y(3+4y)=3(3+4y)\]\
10. Expand: \[5+15y+3y+4y^{2}=9+12y\] ⇒ \[4y^{2}+18y+5=9+12y\] ⇒ \[4y^{2}+6y-4=0\]\
11. Divide by 2: \[2y^{2}+3y-2=0\]. Solve the quadratic: \(D=3^{2}-4\cdot2(-2)=9+16=25\).\
12. \[y=\frac{-3\pm5}{4}\] ⇒ \(y=\frac{2}{4}=\frac12\) or \(y=\frac{-8}{4}=-2\). Since \(0°
0\). Hence \(y=\tan B=\frac12\).\
13. From (iii): \[x=\frac{5}{3+4y}=\frac{5}{3+4\cdot\frac12}=\frac{5}{3+2}=\frac{5}{5}=1\] ⇒ \(\tan A = x = 1\).\
14. Therefore \(A=\tan^{-1}(1)=45^{\circ}\).\
15. \(B=\tan^{-1}\left(\frac12\right)\). In degrees, \(B\approx 26.6^{\circ}\).\
16. Check: \(A+B\approx71.6^{\circ}\) (within the given range) and \(A-B\approx18.4^{\circ}\) with \(\tan(A+B)=3\) and \(\tan(A-B)=\frac13\). The solution satisfies all conditions.