Answer: \(26\) terms; common ratio \(r=\dfrac{\sqrt3}{2}\).

(a) For an arithmetic progression,

\(S_n=\frac n2\left(2a+(n-1)d\right).\)

Here \(a=7\) and \(d=0.4\), so

\(S_n=\frac n2\left(14+0.4(n-1)\right).\)

We need \(S_n\gt300\), so

\(\frac n2\left(14+0.4(n-1)\right)\gt 300.\)

This simplifies to

\(0.4n^2+13.6n-600\gt 0.\)

The positive critical value is approximately

\(25.29.\)

Therefore the least integer value of \(n\) is

\(26.\)

(b) Let the first term be \(a\) and the common ratio be \(r\).

The sum of the first two terms is \(9\), so

\(a+ar=9.\)

The sum to infinity is \(36\), so

\(\frac{a}{1-r}=36.\)

From the second equation,

\(a=36(1-r).\)

Substitute this into \(a(1+r)=9\):

\(36(1-r)(1+r)=9.\)

So

\(36(1-r^2)=9.\)

Hence

\(1-r^2=\frac14.\)

Therefore

\(r^2=\frac34.\)

The terms are positive, so \(r\gt0\). Therefore

\(r=\frac{\sqrt3}{2}.\)