If the curve $y=a x^3+b x^2+c x+5$ touches the x-axis at A(-2, 0) and cuts the y-axis at a point B where its slope is 3. Then,

$a=-\frac{1}{2}, b=-\frac{3}{4}, c=3$

We have,

$y=a x^3+b x^2+c x+5$            ..........(i)

$\Rightarrow \frac{d y}{d x}=3 a x^2+2 b x+c$              ..........(ii)

Since the curve (i) touches x-axis at A(-2, 0). Therefore, point A lies on (i) and $\left(\frac{d y}{d x}\right)_A=0$

∴  $-8 a+4 b-2 c+5=0$           ..........(iii)

and, $12 a-4 b+c=0$          ..........(iv)

The curve (i) cuts y-axis at B(0, 5). It is given that

$\left(\frac{d y}{d x}\right)_B=3 \Rightarrow c=3$

Putting c = 3 in (iii) and (iv) and solving them, we get

$a=-\frac{1}{2}$ and $b=-\frac{3}{4}$