Answer: (a) \(x=4,\ y=16\); (b) \(x=\dfrac12+\dfrac12\ln\dfrac32\).

We start with the main method. Use a substitution such as \(u=a^x\) to turn the exponential equation into a quadratic equation.

For part (a), combine the exponential terms:

\(\mathrm e^{x+y}\times \mathrm e^{3x-2y}=\mathrm e^{4x-y}.\)

Since this is equal to \(1=\mathrm e^0\),

\(4x-y=0.\)

So

\(y=4x.\)

Substitute this into \(x^2y=256\):

\(x^2(4x)=256.\)

Thus

\(4x^3=256,\)

so

\(x=4.\)

Hence

\(y=4x=16.\)

For part (b), let

\(z=\mathrm e^{2x-1}.\)

Then

\(\mathrm e^{1-2x}=\frac1z.\)

The equation becomes

\(10z-11=\frac6z.\)

Multiplying by \(z\),

\(10z^2-11z-6=0.\)

Factorise:

\((5z+2)(2z-3)=0.\)

Since \(z=\mathrm e^{2x-1}\gt 0\), reject \(z=-\frac25\).

So

\(z=\frac32.\)

Therefore

\(\mathrm e^{2x-1}=\frac32.\)

Taking natural logarithms,

\(2x-1=\ln\frac32.\)

Hence

\(x=\frac12+\frac12\ln\frac32.\)

This completes the solution and gives the required result.