Time taken to reach market $t_{1}=\frac{2.5}{5}=0.5$ hour $=30 min$

time taken to get back to home is $t_{2}=\frac {2.5}{7 .5}=.33$hour$=20 min$ 

Average velocity for $0-30$ in is $v=\frac{2.5}{5}=5 km / h$ 

Average speed for $0-30$ in is $s=\frac{2.5}{0.5}=5 km / h$ 

total time he took for travelling $t=30+20=50 min =\frac{5}{6}$ hour

When he reached back then net displacement is zero

so for $0-50$ min

Average velocity for $0-50$ in is $v=\frac{0}{\frac{5}{6}}=0 km / h$

Total distance he traveled when he arrive back is $2.5+2.5=5 km$

Average speed for $0-50$ in is $v=\frac{{5}}{\frac{5}{6}}=6 km / h$