You should have come up with:

[1]

Where N_{0} is one quarter of the initial amount in the pot.
It is l_{p} because there are no biological
eliminations for the pot i.e. l_{b}
= 0

Next, you should be able to get an equation to describe the 360 minute time the isotope was in the owl. The effective half-life in the owl is 4.00 hours, or 240 minutes. Therefore the effective half-life in the owl will be

- eff = 0.693/240
min

[2]

Where N'_{0 }is the amount in both bats at the moment they are
eaten. Therefore 0.5N'_{0} is the final amount in the bat.

Next, you need an equation for the 40 minutes after the bats drank
the coffee and before they are eaten.

[3]

From [2] N'_{0} = 566 {Answer to (a)}

Using this in [3]

N_{0} = 400

So the initial amount in the pot was 4X400 = 1600 counts per minute.
{Answer to (b)}

Now using [1]

l_{p} = (0.693)/400 = 1.73X10^{-3}
min^{-1} {Answer to (c)}

The fourth and final equation you will need relates the three decay constants to each other.

[4] eff = p + b

l_{bB} = l_{e
}-l_{pB }
= (8.66 - 1.73)X10^{-3} min^{-1} {Answer to (d)}