(a) The amount of water lost each day forms an arithmetic progression (AP) with the first term, \(a = 10\), and the common difference, \(d = 2\).

The formula for the nth term of an AP is \(a + (n-1)d\).

For the 30th day, \(n = 30\), so the amount lost is:

\(10 + (30-1) imes 2 = 10 + 29 imes 2 = 10 + 58 = 68\) litres.

(b) The total amount of water lost by the nth day is given by the sum of the first n terms of the AP:

\(S_n = \frac{n}{2} (2a + (n-1)d)\).

We need \(S_n = 2000\):

\(\frac{n}{2} (20 + 2(n-1)) = 2000\)

\(n(20 + 2n - 2) = 4000\)

\(n(18 + 2n) = 4000\)

\(2n^2 + 18n - 4000 = 0\)

Solving this quadratic equation gives \(n = 41\).