Answer: (a) first term \(-16\), common difference \(4\); (b) \(n=26\).

We start with the main method. Identify whether the sequence is arithmetic or geometric, then apply the appropriate formula for the required sum or term.

(a) Let the first term be \(a\) and the common difference be \(d\).

The first three terms are \(a\), \(a+d\), and \(a+2d\). Their sum is \(-36\), so

\(3a+3d=-36.\)

Thus

\(a+d=-12.\)

The last three terms are the 10th, 11th and 12th terms:

\(a+9d,\quad a+10d,\quad a+11d.\)

Their sum is \(72\), so

\(3a+30d=72.\)

Thus

\(a+10d=24.\)

Subtract \(a+d=-12\) from \(a+10d=24\):

\(9d=36,\)

so

\(d=4.\)

Then

\(a+4=-12,\)

so

\(a=-16.\)

(b) The first term is \(1\) and the common ratio is

\(r=\frac{1.2}{1}=1.2.\)

The sum of the first \(n\) terms is

\(S_n=\frac{1(1.2^n-1)}{1.2-1}.\)

So

\(S_n=5(1.2^n-1).\)

We need

\(5(1.2^n-1)\gt500.\)

Therefore

\(1.2^n-1\gt100,\)

so

\(1.2^n\gt101.\)

Taking logarithms,

\(n\log(1.2)\gt\log(101).\)

Hence

\(n\gt\frac{\log(101)}{\log(1.2)}\approx25.3.\)

The smallest integer value of \(n\) is therefore

\(n=26.\)

This completes the solution and gives the required result.