(a) The probability of obtaining a 4 on a single throw is \(\frac{1}{6}\). The probability of not obtaining a 4 is \(\frac{5}{6}\). For the first 7 throws, Ramesh does not get a 4, and on the 8th throw, he gets a 4. The probability is:

\(\left( \frac{5}{6} \right)^7 \times \frac{1}{6} = \frac{78125}{1679616} \approx 0.0465\)

(b) The probability that it takes no more than 5 throws to obtain a 4 is the complement of not obtaining a 4 in all 5 throws:

\(P(X < 6) = 1 - \left( \frac{5}{6} \right)^5\)

Alternatively, it can be calculated as the sum of probabilities of obtaining a 4 on the 1st, 2nd, 3rd, 4th, or 5th throw:

\(\frac{1}{6} + \left( \frac{5}{6} \right) \left( \frac{1}{6} \right) + \left( \frac{5}{6} \right)^2 \left( \frac{1}{6} \right) + \left( \frac{5}{6} \right)^3 \left( \frac{1}{6} \right) + \left( \frac{5}{6} \right)^4 \left( \frac{1}{6} \right)\)

Calculating gives:

\(\frac{4651}{7776} \approx 0.598\)