Let larger tap A takes $$x$$ minutes to fill the tank completely.
Thus, in 1 minute, tap A will fill $$\frac { 1 }{ x } $$ portion of the tank.
By given condition, smaller tap B takes $$x+3$$ minutes to fill tank completely.
Thus, in 1 minute, tap B will fill $$\frac { 1 }{ x+3 } $$ portion of tank
Time taken by both taps to fill tank completely is $$3\frac { 1 }{ 13 }=\frac { 40 }{ 13 } $$ minutes
Thus, in 1 minute, both taps will fill $$=\frac { 13 }{ 40 } $$ portion of the tank.
Thus, we can write,
$$\frac { 1 }{ x } +\frac { 1 }{ x+3 } =\frac { 13 }{ 40 } $$
$$\therefore \frac { x+x+3 }{ x\left( x+3 \right) } =\frac { 13 }{ 40 } $$
$$\therefore \frac { 2x+3 }{ x\left( x+3 \right) } =\frac { 13 }{ 40 } $$
$$\therefore \frac { 2x+3 }{ { x }^{ 2 }+3x } =\frac { 13 }{ 40 } $$
$$\therefore 40\left( 2x+3 \right) =13\left( { x }^{ 2 }+3x \right) $$
$$\therefore 80x+120=13{ x }^{ 2 }+39x$$
$$\therefore 13{ x }^{ 2 }-41x-120=0$$
$$\therefore x=\frac { -\left( -41 \right) \pm \sqrt { { \left( -41 \right) }^{ 2 }-4\times 13\times -120 } }{ 2\times 13 } $$
$$\therefore x=\frac { 41\pm \sqrt { 1681+6240 } }{ 26 } $$
$$\therefore x=\frac { 41\pm \sqrt { 7921 } }{ 26 } $$
$$\therefore x=\frac { 41\pm 89 }{ 26 } $$
We will only consider positive root as time is never negative.
$$\therefore x=\frac { 41+89 }{ 26 } $$
$$\therefore x=\frac { 130 }{ 26 } $$
$$\therefore x=5\quad minutes$$
Thus, larger tap will take $$5\quad minutes$$ to fill the tank completely.
Smaller tap will take $$5+3=8\quad minutes$$ to fill the tank complete