The start time is 3:M where M is the minutes. The finish time is therefore M/5:15 approx where 5<M/5<6, so 25<M<30. That gives us 3:25 as the approx start time and 5:15 as the approx finish time.

However, 25 minutes is 25/60=5/12 of an hour and the hour hand will therefore move 5/12 of the distance between 3 and 4 on the clock face, which is divided into 5 minute divisions. 5/12*5=25/12=2 1/12 minutes. This changes the finish time to 5hr (15+2 1/12)mins=5:17 1/12 or 5:205/12. But, this takes us back to the start time, because 205/12 minutes into the hour means that the hour hand moves 205/(12*60)*5=205/144 minutes after 25 past 3. That revises the start time to 25+205/144=3805/144=26.42 minutes approximately.

Assuming the question only expects answers to the nearest minute then 3:26 should be sufficient for the start time and 5:17 for the finish time. The boy read for 1hr 51 minutes.

(The accurate answer is 3780/143=26.433567 minutes after 3 o'clock is the start time and the finish time is 2460/143=17.202797 minutes after 5 o'clock. The boy read for 1440/13 minutes=1hr 660/13=1hr 50.769231 minutes.

Here's where the figures come from. Let's suppose that x minutes after 3:25 is the true time when the hours and minutes are interchanged. And also suppose that y minutes after 5:15 is the true time, then y=5(25+x)/60=(25+x)/12 and x=(15+y)/12. This is the algebraic equivalent of the arithmetic shown earlier. We have simultaneous equations: 12x=15+y and 12y=25+x. We can write y=12x-15 and substitute: 12(12x-15)=25+x, so 144x-180=25+x, 143x=205, making x=205/143. So the number of minutes after 3 o'clock is 25+205/143=3780/143=26.43 mins approx. And y=12*205/143-15, so 15+y=2460/143=17.20 mins approx.)