$f(x)=y=a x^{3}+b x^{2}+c x+5 \quad \ldots$ (i)

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

Touches $x$-axis at $P(-2,0)$

$\left.\Rightarrow y\right|_{x=-2}=0 \Rightarrow-8 a+4 b-2 c+5=0 \quad \ldots ...(iii)$

Touches $x$-axis at $P(-2,0)$ also implies

$$ \left.\frac{d y}{d x}\right|_{x=-2}=0 \Rightarrow 12 a-4 b+c=0 \quad \ldots...(iv) $$

$y=f(x)$ cuts $y$-axis at $(0,5)$

Given, $\left.\frac{d y}{d x}\right|_{x=0}=c=3$

From (iii), (iv) and (v)

$$ \begin{aligned} &a=-\frac{1}{2}, b=-\frac{3}{4}, c=3 \\\\ &\Rightarrow f(x)=\frac{-x^{2}}{2}-\frac{3}{4} x^{2}+3 x+5 \\\\ &f^{\prime}(x)=\frac{-3}{2} x^{2}-\frac{3}{2} x+3 \\\\ &=\frac{-3}{2}(x+2)(x-1) \\\\ &f^{\prime}(x)=0 \text { at } x=-2 \text { and } x=1 \end{aligned} $$

By first derivative test $x=1$ in point of local maximum Hence local maximum value of $f(x)$ is $f(1)$

i.e., $\frac{27}{4}$