Answer: \(A\left(-\dfrac52,0\right)\), \(B\left(\dfrac12,-6\right)\); area \(=9+\ln8\).

(a) At \(A\), the line meets the \(x\)-axis, so \(y=0\). From

\(2x+y=-5,\)

we get

\(2x=-5,\)

so

\(A\left(-\frac52,0\right).\)

For \(B\), use \(y=-5-2x\) and \(xy+3=0\). Substituting gives

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

So

\(2x^2+5x-3=0.\)

Factorising,

\((2x-1)(x+3)=0.\)

The point \(B\) shown on the diagram is on the right-hand branch of the curve, so \(x=\frac12\). Then

\(y=-5-2\left(\frac12\right)=-6.\)

Therefore

\(B\left(\frac12,-6\right).\)

(b) The shaded region can be split into a triangle and a region under the curve.

The triangle has base

\(\frac52+\frac12=3\)

and height \(6\), so its area is

\(\frac12(3)(6)=9.\)

For the curved part, the curve is

\(y=-\frac3x.\)

The area between the curve and the \(x\)-axis from \(x=\frac12\) to \(x=1\) is

\(\int_{\frac12}^{1}\left(0-\left(-\frac3x\right)\right)\,dx =\int_{\frac12}^{1}\frac3x\,dx.\)

So this area is

\(\left[3\ln x\right]_{\frac12}^{1} =3\ln1-3\ln\frac12 =3\ln2 =\ln8.\)

Therefore the total area is

\(9+\ln8.\)