Answer: (a) first term \(3\), common difference \(2\); (b) common difference \(9\), common ratio \(2\).

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 5th term is \(a+4d\), so

\(a+4d=11.\)

The 7th term is three times the 2nd term, so

\(a+6d=3(a+d).\)

Simplify the second equation:

\(a+6d=3a+3d,\)

so

\(3d=2a.\)

Hence

\(a=\frac32d.\)

Substitute into \(a+4d=11\):

\(\frac32d+4d=11.\)

Thus

\(\frac{11}{2}d=11,\)

so

\(d=2.\)

Therefore

\(a=3.\)

(b) Let the common difference of the arithmetic progression be \(d\), and let the common ratio of the geometric progression be \(r\).

The arithmetic progression begins

\(3,\ 3+d,\ 3+2d,\ldots\)

and the geometric progression begins

\(3,\ 3r,\ 3r^2,\ 3r^3,\ 3r^4,\ldots\)

The 2nd term of the arithmetic progression is equal to the 3rd term of the geometric progression:

\(3+d=3r^2.\)

The 6th term of the arithmetic progression is equal to the 5th term of the geometric progression:

\(3+5d=3r^4.\)

From the first equation,

\(d=3r^2-3.\)

Substitute this into the second equation:

\(3+5(3r^2-3)=3r^4.\)

Simplify:

\(15r^2-12=3r^4.\)

Divide by \(3\):

\(5r^2-4=r^4.\)

So

\(r^4-5r^2+4=0.\)

Factorise:

\((r^2-1)(r^2-4)=0.\)

Since the common ratio is greater than \(1\),

\(r=2.\)

Then

\(d=3(2^2)-3=9.\)

Therefore the common difference is \(9\), and the common ratio is \(2\).

This completes the solution and gives the required result.