(a) To find \(a\), calculate the probability of obtaining exactly 1 head:

\(a = 0.7 \times (0.5)^3 + 0.3 \times (0.5)^3 = \frac{1}{5}\).

For \(b\), calculate the probability of obtaining exactly 2 heads:

\(b = 0.7 \times 0.5^3 \times 3 + 0.3 \times 0.5^3 \times 3 = \frac{3}{8}\).

For \(c\), calculate the probability of obtaining exactly 3 heads:

\(c = 0.7 \times 0.5^3 \times 3 + 0.3 \times 0.5^3 = \frac{3}{10}\).

(b) The expected value \(E(X)\) is calculated as:

\(E(X) = \frac{3 \times 0 + 16 \times 1 + 30 \times 2 + 24 \times 3 + 7 \times 4}{80} = \frac{176}{80} = 2.2\).

(c) The probability of obtaining exactly one head on fewer than 3 occasions is:

\(P(0, 1, 2) = \binom{10}{0} 0.2^0 0.8^{10} + \binom{10}{1} 0.2^1 0.8^9 + \binom{10}{2} 0.2^2 0.8^8\).

Calculating gives \(0.107374 + 0.268435 + 0.301989 = 0.678\).

(d) The probability that Jacob obtains exactly one head for the first time on the 7th or 8th time is:

\(0.8^6 \times 0.2 + 0.8^7 \times 0.2 = 0.0524288 + 0.041943 = 0.09437\).