Answer: (a) \(x=16,\ y=64\); (b) \(t=2\).

We start with the main method. Use logarithm laws and index laws first so that the equation is reduced to an algebraic equation.

(a) Let

\(A=\log_2x,\quad B=\log_2y.\)

The equations become

\(3A+2B=24\)

and

\(5A-3B=2.\)

Multiply the first equation by \(3\) and the second by \(2\):

\(9A+6B=72,\)

\(10A-6B=4.\)

Adding gives

\(19A=76,\)

so

\(A=4.\)

Substitute into \(3A+2B=24\):

\(12+2B=24,\)

so

\(B=6.\)

Therefore

\(\log_2x=4,\quad \log_2y=6,\)

and hence

\(x=16,\quad y=64.\)

(b) Use the laws of indices:

\(\frac{2^{t+4}}{2^{1-2t}}=2^{t+4-(1-2t)}=2^{3t+3}.\)

Since \(512=2^9\),

\(2^{3t+3}=2^9.\)

Thus

\(3t+3=9,\)

so

\(t=2.\)

This completes the solution and gives the required result.