1. Use continuity of $f(x)$ at $x=0$

Given

Since both pieces are individually continuous, for $f$ to be continuous on $\mathbb R$, it must be continuous at $x=0$.

• Left value at $x=0$: $f(0)=0+a=a$
• Right-hand limit at $x\to 0^+$: $|0-4|=4$

So, $a=4$

Hence,

2. Use continuity of $g(x)$ at $x=0$

Given

Again, continuity is needed only at $x=0$.

• Left-hand limit at $x\to 0^-$: $0+1=1$
• Value at $x=0$ from the right piece: $g(0)=(0-4)^2+b=16+b$

For continuity, $16+b=1$ $b=-15$

Thus,

3. Compute $(g\circ f)(2)$

First find $f(2)$. Since $2>0$, use $f(x)=|x-4|$: $f(2)=|2-4|=2$

Now find $g(2)$. Since $2\ge 0$, use $g(x)=(x-4)^2-15$: $g(2)=(2-4)^2-15=4-15=-11$

Therefore, $(g\circ f)(2)=g(f(2))=g(2)=-11$

4. Compute $(f\circ g)(-2)$

First find $g(-2)$. Since $-2<0$, use $g(x)=x+1$: $g(-2)=-2+1=-1$

Now find $f(-1)$. Since $-1\le 0$, use $f(x)=x+4$: $f(-1)=-1+4=3$

Therefore, $(f\circ g)(-2)=f(g(-2))=f(-1)=3$

5. Add the two values

So the required value is $\boxed{-8}$

This corresponds to Option D.