Answer: \((1,-4)\), \((-1,4)\), \((2,-2)\), \((-2,2)\).

The key point is to reduce the equation to a standard trigonometric equation, then use the given interval to list all valid solutions.

From

\(xy+4=0,\)

we have

\(xy=-4.\)

Substitute this into the first equation:

\(4x^2+3xy+y^2=8.\)

This gives

\(4x^2+3(-4)+y^2=8.\)

So

\(4x^2+y^2=20. \tag{1}\)

Since \(y=-\frac4x\), substitute into (1):

\(4x^2+\frac{16}{x^2}=20.\)

Multiply by \(x^2\):

\(4x^4-20x^2+16=0.\)

Divide by \(4\):

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

Factorise:

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

Thus

\(x=\pm1,\quad \pm2.\)

Using \(y=-\frac4x\), the solutions are

\((1,-4),\quad(-1,4),\quad(2,-2),\quad(-2,2).\)