Answer: (a) \(1024+15360x+103680x^2\). (b) \(51963120\).

(a) Since

\(9x^2+12x+4=(3x+2)^2,\)

we have

\((9x^2+12x+4)^5=(3x+2)^{10}.\)

The first three terms in ascending powers of \(x\) are

\(2^{10}+{}^{10}C_1(2^9)(3x)+{}^{10}C_2(2^8)(3x)^2.\)

Thus

\(1024+15360x+103680x^2.\)

(b) The general term is

\({}^{12}C_r\left(\frac{6}{x^2}\right)^{12-r}\left(\frac{x^4}{2}\right)^r.\)

The power of \(x\) is

\(-2(12-r)+4r=-24+6r.\)

For the term independent of \(x\), set \(-24+6r=0\), giving \(r=4\).

The required term is

\({}^{12}C_4\left(\frac{6}{x^2}\right)^8\left(\frac{x^4}{2}\right)^4=51963120.\)