About climb: If we use T/W as input, the second grade equation yields two solutions for W/S. Are both solutions realistic?
Raymer's G is a little unusual. In effect, it is the tangent of the flight path angle $\gamma$, and since his equations are for small values of $\gamma$, we can assume that it is approximately equal to $sin\gamma$ or $\gamma$ itself when expressed in radians.
$$G \approx \gamma$$
Next, the expression $\pi\cdot AR\cdot e$: This is part of the square root of the lift coefficient for minimum drag. The complete equation is:
$$c_{L\:(minimum\:drag)} = \sqrt{c_{D0}\cdot\pi\cdot AR\cdot e} = c_{L\:(md)}$$
Next, I need to express that thrust-to-weight ratio differently. As long as we keep $\gamma$ small:
$$\gamma = \frac{T}{W} - \frac{c_D}{c_L} =&gt; \frac{T}{W}-\gamma = \frac{c_D}{c_L}$$
Now I can rewrite the equation of your first question:
$$\frac{W}{S} = q\cdot\frac{\frac{c_D}{c_L}\cdot \frac{c^2_{L\:(md)}}{c_{D0}}}{2} ± q\cdot\frac{\sqrt {\left(\frac{c_D}{c_L}\right)^2\cdot\left(\frac{c^2_{L\:(md)}}{c_{D0}}\right)^2 - 4\cdot c^2_{L\:(md)}}}{2}$$
and when I restrict the equation to the minimum drag point where $c_{D0} = c_{Di}$, I can write
$$\frac{W}{S} = q\cdot c_{L\:(md)} ± q\cdot\sqrt {c^2_{L\:(md)} - c^2_{L\:(md)}}$$
Fancy that! The expression under the root becomes zero when I look at the polar point of minimum drag! The only variable term under the square root is the ratio $\frac{c_D}{c_L}$, and it reaches its minimum at the $c_{L\:(md)}$ point. When I move away from this polar point, I should get two valid solutions.

I guess these two wing loadings allow to fly with the specified flight path angle at the given polar point and thrust-to-weight ratio. Since the weight is prescribed by the thrust-to-weight ratio, the solution is actually for two different wing areas. The smaller wing requires a higher speed and the larger wing a smaller speed than what you get at the lowest drag point. Between both, the speed and wing area should result in configurations which achieve a higher flight path angle $\gamma$, and outside of this interval no level flight should be achievable with the resulting wing areas and speeds.

I cannot follow how you arrive at the second question. The term under the square root should be
$$\left(\frac{T}{W}\right)^2 - 2\cdot\frac{T}{W}\cdot G + G^2 ≥ \frac{4\cdot c_{D0}}{\pi\cdot AR\cdot e}$$
But when I insert again as explained above, I can write for the term under the root:
$$\left(\frac{c_D}{c_L}\right)^2 - \frac{4\cdot c^2_{D0}}{c^2_{L\:(md)}}$$
so the condition for a positive root is
$$\frac{c_{D0}+c_{Di}}{c_L} ≥ \frac{2\cdot c_{D0}}{c_{L\:(md)}}$$
or
$$ \frac{c_{D0}}{c_L} + \frac{c_L}{\pi\cdot AR\cdot e} ≥ 2\cdot\sqrt{ \frac{c_{D0}}{\pi\cdot AR\cdot e}}$$
which is certainly true; both sides of the equation are equal at the lowest drag polar point and away from it the left side is bigger and the right side stays constant.

If you just isolate the term under the root and look at it by itself, I doubt you have something that corresponds to a realistic configuration. You need to extract both numerator and denominator to keep the physical analogies intact. What Raymer does is to look at the term under the root in isolation and then postulates that the first summand must be larger than the second. He puts both on the two sides of an inequality and extracts the roots on both sides separately. Then he moves G over to the other side. There is no second version of the inequality possible.