1. Since the curve $y(x)=ax^3+bx^2+cx+5$ touches the $x$-axis at $P(-2,0)$, we have:

• $y(-2)=0$
• $y'(-2)=0$ (because touching the $x$-axis means the tangent there is horizontal)

Also, it cuts the $y$-axis at $Q$, where $y'=3$. Since the $y$-axis corresponds to $x=0$, this gives: $y'(0)=3$

2. Compute the derivative: $y'(x)=3ax^2+2bx+c$

From $y'(0)=3$, we get: $c=3$

3. Use $y(-2)=0$: $a(-2)^3+b(-2)^2+c(-2)+5=0$ $-8a+4b-2c+5=0$ Substitute $c=3$: $-8a+4b-6+5=0$ $-8a+4b-1=0$ $4b-8a=1 \quad (1)$

4. Use $y'(-2)=0$: $3a(-2)^2+2b(-2)+c=0$ $12a-4b+c=0$ Substitute $c=3$: $12a-4b+3=0$ $12a-4b=-3 \quad (2)$

5. Solve equations (1) and (2): $(1)\; 4b-8a=1$ $(2)\; 12a-4b=-3$

Add them: $4a=-2$ $a=-\frac12$

Now from (1): $4b-8\left(-\frac12\right)=1$ $4b+4=1$ $4b=-3$ $b=-\frac34$

Thus, $a=-\frac12,\quad b=-\frac34,\quad c=3$

So, $y(x)=-\frac12x^3-\frac34x^2+3x+5$

6. Find critical points: $y'(x)=3ax^2+2bx+c=-\frac32x^2-\frac32x+3$ Factor: $y'(x)=-\frac32(x^2+x-2)$ $y'(x)=-\frac32(x+2)(x-1)$

Hence critical points are: $x=-2,\;1$

7. Determine which is the local maximum using second derivative: $y''(x)=6ax+2b=-3x-\frac32$

At $x=-2$: $y''(-2)=6-\frac32=\frac92>0$ So $x=-2$ is a local minimum.

At $x=1$: $y''(1)=-3-\frac32=-\frac92<0$ So $x=1$ is a local maximum.

8. Compute the local maximum value: $y(1)=-\frac12-\frac34+3+5$ $=8-\frac12-\frac34$ $=8-\frac54$ $=\frac{32-5}{4}=\frac{27}{4}$

9. Compare with options: $\frac{27}{4}$ is option A.

So the local maximum value is: $\boxed{\frac{27}{4}}$