Note the rankings, figures and methods given here are not definitive - they are only meant to provide an example of how to approach the problem. The important thing is that any estimates have been justified.
Energy
(Highest) 12 4 3 (Lowest), although this will depend on the mass of the person in question.
1) Mass of a typical person $ m = 60 \textrm{ kg}$
Height travelled up $h = 818 \textrm{ m}$
$\therefore$ assuming other energy uses (overcoming friction etc.) are insignificant in comparison to gravitational potential
$ E_1 \approx mgh \approx 60 \times 9.8 \times 818 \approx 500 \textrm{ kJ}$
2) Energy in can of full-sugar coke (from nutritional information) is about $180 \textrm{ kJ} $ per $100$ ml. A can contains $330$ ml of drink. Thus:
$E_2 \approx 594 \textrm{ kJ} $.
3) 1 atom of lead has mass $m = \frac{m_{Pb}}{N_A} $
Einstein's equation $E = mc^2$
$E_2 = \left(\frac{m_{Pb}}{N_A}\right) c^2 = \left(\frac{207}{6.022 \times 10^{23}}\right) \times (3.00 \times 10^8)^2 \approx 3 \times 10^{-5} \textrm{ J}$
4) Change in internal energy due to raising temperature of mass $m$ of water by $\Delta T$ $\Delta U = mc_p\Delta T$
Specific heat capacity of water (at constant pressure) $c_p \approx 4.2 \textrm{ kJ (kg K)}^{-1}$
Change in temperature from room temperature ($\approx 25\ ^\circ C$) to boiling ($= 100\ ^\circ C$) $\Delta T = 75 \textrm{ K}$
Typical volume of water to be boiled $\approx 1 \textrm{ dm}^{-3}$ which has mass $m = 1 \textrm{ kg}$
$E_4 \approx 1 \times 4200 \times 75 \approx 300 \textrm{ kJ}$
Time
(Longest) 4 1 3 2 (Shortest)
1) Circumference of Earth $C \approx 40\ 000 \textrm{ km}$
Speed of light $c \approx 3 \times 10^8 \textrm{ ms}^{-1}$
$t_1 = \frac{C}{c} \approx \frac{40 \times 10^6}{2 \times 3 \times 10^8} \approx 0.07 \textrm{ s}$
2) Top sprinter can complete 100 m in 9 s
Assuming constant speed (actually varies substantially however sufficient for this calculation) $v \approx 11 \textrm{ ms}^{-1}$
To travel $1 \textrm{ mm}$ therefore would take
$t_2 \approx \frac{1 \times 10^{-3}}{11} = 9 \times 10^{-5}$
3) Angular speed of second hand $\omega = \frac{2\pi}{60} = \frac{pi}{30} \textrm{ rad s}^{-1}$
Typical length of second hand $\ell = 1 \textrm{ cm}$
Therefore speed of second hand $v = \ell \omega = \frac{2\pi}{6000} \textrm{ ms}^{-1}$
Therefore time taken to travel 1 micron
$t_3 \approx \frac{1 \times 10^{-6} \times 6000}{2\pi} \approx 1 \times 10^{-3} \textrm{ s}$
4) Flame front will travel at flame speed of hydrogen $v \approx 4 \textrm{ ms}^{-1}$ ??
Typical length of a test tube $\ell \approx 15 \textrm{ cm}$
$t_4 \approx \frac{0.15}{4} \approx 4 \times 10^{-2} \textrm{ s}$
Distance
(Longest) 3 2 4 1 (Shortest)
1) Gravitational field of Moon $\approx \frac{1}{6}$ of Earth's
Typical height centre of mass is raised through when jumping on Earth $h = 0.5 \textrm{ m}$
$\therefore$ height reached on Moon
$d_1 \approx 6 \times 0.5 = 3 \textrm{ m}$
2) Density of lead $\rho \approx 11 \textrm{ g cm}^{-3}$
Radius of tennis ball $r \approx 3.5 \textrm{ cm}$
Volume of tennis ball $V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi \times (3.5 \times 10^{-2})^3 \approx 180 \textrm{ cm}^3$
$\therefore$ tennis ball sized lump of lead has mass $m = \rho V = \frac{11 \times 180}{100} \approx 2 \textrm{ kg}$
Mass of shot put 4-7 kg, typical range of trained atheletes being $\geq 15 \textrm{ m}$ (WR 23.12 m)
$\therefore$ for an object of half the mass assume an average person could reach similar ranges
$d_2 \approx 15 \textrm{ m}$
3) From experimentation, maximum clapping frequency approximately $f \approx 10 \textrm{ Hz}$
$\therefore$ time between each clap $T \approx 0.1 \textrm{ s}$
Speed of sound $v \approx 330 \textrm{ ms}^{-1}$
$d_3 \approx 33 \textrm{ m}$
4) Typical time to sprint 100 m for an untrained person is around 15 s
Therefore in 1 s cover around $\frac{100}{15} \textrm{ m}$
$d_4 \approx 7 \textrm{ m}$
Mass
(Highest) (3 1 4) 2 (Lowest)
Large degree of uncertainty in top three
1) Air pressure at sea level $p \approx 10^5 \textrm{ Pa}$
Surface area of Earth $A \approx 510 \times 10^9 \textrm{ m}^2$
$\therefore$ weight of atmosphere $W \approx 5.1 \times 10^{16} \textrm{ N}$
Acceleration due to gravity $g \approx 9.8 \textrm{ ms}^{-1}$
$m_1 \approx \frac{W}{g} = 5.2 \times 10^{15} \textrm{ kg}$
2) Average mass of one person (all ages and genders) $m \approx 60 \textrm{ kg}$
Number of people on Earth $P \approx 6 \times 10^9$
$m_2 \approx Pm = 3.6 \times 10^{11} \textrm{ kg}$
3) Density of ice $\rho \approx 920 \textrm{ kg m}^{-3}$
Area of north polar ice cap $A \approx 10^{13} \textrm{ m}$
Typical depth of ice $d \approx 2 \textrm{ m}$
$m_3 \approx \rho Ad = 1.8 \times 10^{16} \textrm{ kg}$
4) Number of living bacteria on Earth $N \approx 10^{30}$
Mass of single bacteria $m \approx 10^{-15} \textrm { kg}$
$m_4 \approx Nm = 10^{15} \textrm{ kg}$