The "three solutions" referred to are three distinct values of $x$ that make the equation true. As you will see later, one of those values must be a repeated root, so arguably there are four roots, but let that go for the moment...
This equation can be restated as $|x^2+4x+3|=m(x-2)$. Writing it this way, it becomes clear that we are looking for points of intersection of the line $y=m(x-2)$ and the curve $y=|x^2+4x+3|$.
The curve is large and positive for large negative values of $x$, decreases to 0 at $x=-3$, rises to a maximum at $x=-2$ and falls to ) at $x=-1$ before rising again to become large and positive as $x$ becomes large and positive. This pattern can be described as a valley with a hill on the valley floor.
The line must pass through the point $(2,0)$, but its gradient depends on $m$. Lines with a positive gradient will never intersect the curve, so the gradient must be negative.
There are three possible types of intersection:
1) The line does not intersect the "hill", so intersects the curve exactly twice (at the valley walls). This means the equation would have exactly two roots.
2) The line intersects the "hill" twice as well as the valley walls. This means the equation would have exactly four roots.
3) The line is a tangent to the "hill" as well as intersecting the valley walls. This means the equation would have exactly three roots.
It is the third scenario that you need to find.
This is the "Type 2" that you have referred to in your question.
The equation is $-x^2-4x-3=m(x-2)$
$-x^2-4x-3=mx-2m$
$-x^2-4x-mx-3+2m=0$
$x^2+(4+m)x+(3-2m)=0$
Repeated root means $(4+m)^2-4(3-2m)=0$
Like you I get $m=-8\pm2\sqrt{15}$.
The very steep root can be discounted because it will not work.
So $m=-8+2\sqrt{15}$.
Best Answer
Well, let's test them out.
$$||x-a|-3|=3$$
The left-hand side before the final modulus is applied must be either $\pm 3$ i.e. $|x-a|-3=3$ or $|x-a|-3=-3$. In the former case, $|x-a|=6$ will have the solutions $x=a+6$ and $x=-6+a$ (can you see why?). In the latter case, we get $|x-a|=0$ so $x=a$ is the unique solution.
That's three solutions.
Now let's try: $$||x-a|+3|=3$$
Either $|x-a|+3=3$ or $|x-a|+3=-3$. In the former case, $|x-a|=0$ so $x=a$. In the latter case, $|x-a|=-6$. This has no solutions (why?). So there is only one solution.
Thus, it is $b=3$ that we want.
This may also give you some insight into how you could have found the solutions without extra candidate solutions. How could we reason by cases about $||x-a|-3|=3$?