Can someone please help me with part b and c.
to a)
Plug in D = 10 and t = 5. Use a calculator. I've got $\displaystyle x \approx 5.353\ \rm{mg}$
to b)
Now D = 10 + 5.353 and t = 1. Use a calculator. I've got $\displaystyle x \approx 13.54897\ \rm{mg}$
to c)
Solve the equation for t:
$\displaystyle 3 = 15.353\cdot e^{-\frac18 t}~\implies~\dfrac{3}{15.353} = e^{-\frac18 t}$
Use logarithm:
$\displaystyle \ln\left(\dfrac{3}{15.353} \right) = -\frac18 t$
I'll leave the rest for you.
Spoiler: