Okay I think this is it, I may be wrong i)
after 0 years t=0
so if you substitute 0 for t in the equation you get the original mass of the rock
m=400+80e^0
so m=480
so when t=10 the mass= 99.8% of 480
= 479.04
so the equation is 479.04= 400+80e^(-k10)
79.04=80e^(-k10)
0.988=e^(-k10)
ln0.988=-k10
(ln0.988)/(-10)=k
(0.001) 3dp=k
this gives the equation m=400+ 80e^(-0.001t)
use this to answer 2 and 3 by substitution