Let's solve this problem together! Given that \(5^{(x + 2y)} = 5\) and \(4^{(x + 3y)} = 16\), we can use these equations to find the values of x and y.

Since \(5^{(x + 2y)} = 5\), we can take the logarithm of both sides with base 5 to get:

x + 2y = log₅5
=> x + 2y = 1

Similarly, since \(4^{(x + 3y)} = 16\), we can take the logarithm of both sides with base 4 to get:

x + 3y = log₄16

=> x + 3y = 2

Now, we have a system of two equations with two unknowns. We can solve this system by subtracting the first equation from the second equation to get:

x + 3y - (x + 2y) = 2 - 1
=> y = 1

Substituting y = 1 into the first equation gives:

x + 2y = 1

=> x + 2(1) = 1

=> x = -1

Now that we know the values of x and y, we can use these values to find the value of \(3^{(x + y)}\):

\(3^{(x + y)}\) = \(3^{(-1 + 1)}\) = \(3^0\) = 1