$Here, the time taken by the train to pass the man is equal to the time taken to pass it\text{'}s length 150 meters.$

$\therefore The speed of the train=\frac{Dis\tan ce covered by the train }{time}=\frac{150}{3}m{s}^{-1}=\frac{150}{3}\times \frac{18}{5}=180kmph$

$Now, relative speed = Speed of train - Speed of man$

$=180 Speed of train - 2$

$\therefore Speed of train = 180+2= 182 kmph.$

$$\begin{gathered} Sam runs ahead of Lisa in 15 seconds = 5 x 15=75 m \\ Here, Lisa\text{'}s average speed = 8m{s}^{-1} \\ And Sam\text{'}s average speed = 5m{s}^{-1} \\ \therefore Lisa\text{'}s speed is=8-5=3m{s}^{-1} more than that of Sam \end{gathered}$$

$\therefore Time taken by Lisa to catch up Sam$

$=\frac{75m}{3 m{s}^{-1}} [As,t=\frac{d}{v}]$

$= 25 seconds$

$\therefore Total time taken by Lisa to catch sum up$

$=25+15=40 seconds$

$Here, each day Peter can upload$

$= 5 \times 60 \times 60 \times 7 = 126000 kb$

$=\frac{126000}{1024}mb=123.05 mb [Given, 5 hours a day; I hour = (60 x 60) seconds]$

$\therefore Maximum number of images that can be uploaded=\frac{123.05}{15}=8.20=8$

$Total value of initial configeration = 13 hamper \times \$12 = \$156$

$Again total value of new configeration = 12 hamper \times \$13=\$156$

$\therefore$The value of gift certificates will not be changed.

$$\begin{gathered} Now Susan started with 13 hampers. \\ Susan has 13 piece of \$3 worth gift \end{gathered}$$

$$\begin{gathered} \therefore Susan has 13 piece of \$4 worth gift \\ \therefore Susan has 13 piece of \$5 worth gift \\ \therefore Possible set \{5, 5, 3\} \{4, 4, 5\} , (3, 3, 3, 4) in which \$5 contains in first two of them and the total \end{gathered}$$

number is 13.

Now, let Susan creates

x hampers of {3, 3, 3, 4}

y hampers of {4,4,5}

And z hampers of {5, 5, 3}

Number of $3 gift will be, 3x + z =13.....(i)

$\therefore Number of \$4 gift will be, x + 2y =13.....\left(ii\right)$

$$\begin{gathered} \therefore Number of \$5 gift will be, y + 2z = 13 \\ Now, multiplying equation \left(iii\right) by 2 and subtracting \left(ii\right) from \left(iii\right) we get , \end{gathered}$$

$$\begin{gathered} \frac{2y+4z=26 \\ x+2y=13 \\ - - - }{4z-x=13...............\left(iv\right)} \end{gathered}$$

$$\begin{gathered} Again, multiplying equation \left(i\right) by 4 and subtracting equation \left(iv\right) from \left(i\right) we get \\ \frac{12x+4z=52 \\ 4z-x=13}{13x=39} \\ \therefore x=3 \\ Here, x is the number of gift hampers without \$5 certificates. \\ \therefore 3 gift hampers did not contain \$5 gift certificates. \end{gathered}$$

$$\begin{gathered} As per divisors and remainders, the difference are as follows \\ In first case, 44-31=13 \\ In second case, 56-43= 13 \\ In third case, 32-19=13 \\ Now, the L..C.M of 44, 56 and 32 is 2464 \\ \therefore The required least number = 2464-13=2451 \end{gathered}$$