1. Use the given condition

We are given $\sin x + \sin^2 x = 1, \qquad x \in \left(0,\frac{\pi}{2}\right).$

Let $s = \sin x.$ Then $s + s^2 = 1 \implies s^2 + s - 1 = 0.$

So, $s = \frac{-1 \pm \sqrt{5}}{2}.$

Since $x \in \left(0,\frac{\pi}{2}\right)$, we have $\sin x>0$, hence $s = \frac{\sqrt{5}-1}{2}.$

Also, from $1-s^2=s,$ we get $\cos^2 x = s.$

Now, $\tan^2 x = \frac{\sin^2 x}{\cos^2 x} = \frac{s^2}{s} = s.$

Thus, $\cos^2 x = \tan^2 x = s.$

2. Convert all powers in terms of $s$

Since $\cos^2 x = s$ and $\tan^2 x = s$, $\cos^{12}x = (\cos^2 x)^6 = s^6, \qquad \tan^{12}x = (\tan^2 x)^6 = s^6,$ $\cos^{10}x = s^5, \qquad \tan^{10}x = s^5,$ $\cos^8x = s^4, \qquad \tan^8x = s^4,$ $\cos^6x = s^3, \qquad \tan^6x = s^3.$

Therefore the given expression becomes $E = (s^6+s^6)+3(s^5+s^5+s^4+s^4)+(s^3+s^3).$

So, $E = 2s^6 + 6s^5 + 6s^4 + 2s^3$ $= 2s^3(s^3+3s^2+3s+1)$ $= 2s^3(s+1)^3.$

3. Use the relation $s^2+s=1$

From $s^2+s=1,$ we have $s(s+1)=1.$

Hence, $s^3(s+1)^3 = [s(s+1)]^3 = 1^3=1.$

Therefore, $E = 2\cdot 1 = 2.$

4. Check options

The value is $\boxed{2}.$

So the correct option is C.