Let's narrow the three possibilities to one, and confirm whether it actually works.
First, note that there are three possibilities for a graph of $y=x^2+bx+c$: It passes above the the $x$-axis (and has no real roots), it just 'kisses' the $x$-axis (and has one real root), or it crosses the axis twice (two real roots). In the first two cases, the quadratic is always positive and so the absolute value has no effect. But, in the third case, the portion of the parabola below the $y$-axis is reflected into the $x>0$ half plane.
Suppose we now cross these graphs with a horizontal line $y=k$. If we have a parabola, then this line can cross the parabola at most twice; since we want four, we must have the two real roots of $x^2+bx+c=0$. If we recall the quadratic formula, this requires that $b^2-4c>0$ so that the plus/minus gives two real solutions. That eliminates option (b).
If we do have $b^2-4c>0$, then we can choose a range of $k$ such that it crosses the reflected parabola twice. By visual inspection of the reflected parabola, it should be of the form $0<k<k_{max}$. Only option (a) is of this form, with $k_{max}=\frac{4c-b^2}{4}$. To check this, note that the most negative value that $x^2+bx+c=(x+\frac{b}{2})^2+\frac{4c-b^2}{4}$ can express is $\frac{4c-b^2}{4}$ (which is negative since we require $b^2-4c>0$.) Upon taking the absolute value, this means that the reflected 'hump' of the parabola has a height of $\frac{4c-b^2}{4}$ which is indeed $k_{max}$ as in answer (a).
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
For $k\le 0$, there will be no real solution at all for $kx^2\le 0<e^x$.
For $k>0$ there will always be exactly one solution with $x\le 0$. For small positive $k$, there is no positive solution, for suitable $k$ the curves will touch and for larger $k$ we have two positive real solutions (so three real solutions in total). We ned to find the boundary case that the curves touch, i.e., find $k>0$ and $x_0>0$ such that the functions and derivative agree at $x_0$: $$ e^{x_0}=kx_0^2\qquad \text{and}\qquad e^{x_0}=2kx_0$$ We conclude $x_0=2$, hence $$\tag1k=\frac{e^2}4.$$ The value given by $(1)$ is the infimum of all $k$ for which there are exactly three distinct real solutions. If we count the tangential case as a solutoin of multiplicity two, then the value given by $(1)$ is also the minimum value of $k$ forwhich there are (with multiplicity) three real solutions.