(i) To find the first 3 terms in the expansion of \((2 + x^2)^5\), we use the binomial theorem:

\((2 + x^2)^5 = \binom{5}{0} 2^5 + \binom{5}{1} 2^4 x^2 + \binom{5}{2} 2^3 (x^2)^2 + \ldots\)

Calculating each term:

First term: \(\binom{5}{0} 2^5 = 32\)

Second term: \(\binom{5}{1} 2^4 x^2 = 5 \times 16 x^2 = 80x^2\)

Third term: \(\binom{5}{2} 2^3 (x^2)^2 = 10 \times 8 x^4 = 80x^4\)

Thus, the first three terms are \(32 + 80x^2 + 80x^4\).

(ii) Now, expand \((1 + x^2)^2\):

\((1 + x^2)^2 = 1 + 2x^2 + x^4\)

To find the coefficient of \(x^4\) in \((1 + x^2)^2(2 + x^2)^5\), consider the product of terms that result in \(x^4\):

- \(1 \times 80x^4 = 80x^4\)

- \(2x^2 \times 80x^2 = 160x^4\)

- \(x^4 \times 32 = 32x^4\)

Adding these, the coefficient of \(x^4\) is \(80 + 160 + 32 = 272\).