Answer
5% more work will be done by engine after tuning.
Work Step by Step
Suppose before tuning the engine efficiency was $e_1$
Heat input is given as $Q_1=1300J$
suppose the work done is $W_1$
from definition of efficiency of engine $e_1=\frac{W_1}{Q_1}$
So $W_1=e_{1}Q_{1}$
$W_1=1300e_{1}J$.....................(1)
suppose after tuning the engine the efficiency is $e_2$
given that new efficiency is 5% more than old efficiency $e_2=e_1 +\frac{5e_1}{100}$
suppose in this case work done by the engine is $W_2$
Heat input is same as $Q_1=1300J$
from definition of efficiency of engine $e_2=\frac{W_2}{Q_1}$
so work done by the engine after tuning is $W_2=e_{2}Q_{1}$
so $W_2=(e_1 +\frac{5e_1}{100})1300J$................equation(2)
difference in work done after tuning
increase in work done
$\Delta W=W_2-W_1$
$\Delta W=(e_1 +\frac{5e_1}{100})1300J-1300e_{1}J$
amount of work increased
$\Delta W=1300J\times\frac{5e_1}{100}$
so % increase in work $\Delta W$%=$\frac{\Delta W}{W_1}\times100$
$\Delta W$%=$\frac{1300J\times\frac{5e_1}{100}}{1300e_{1}J}\times100$ %
$\Delta W$%=$5$%
so 5% more work will be done by engine after tuning.