I do not understand your calculations. (You are working partly in kgwt and partly in Newtons, which is confusing. Stick to one unit and use throughout.)

I would start from the outside and work inwards :

There are 3 distinct joints (nodes) to examine. At each of them there are 3 forces in balance.

The tension in the 4 longitudinal cable ends can be found because the vertical components of this tension supports the whole weight of the bridge ($4.80\times 10^4 kg$).

The 12 vertical cables each carry the same weight, which in total is again the full weight of the bridge. Balancing forces at joint B allows you to find the force and angle of cable BA.

Repeat the same process for joint A to find the tension and angle of cable AC where C is the innermost joint/node.

The lengths of the vertical cables can be found from the angles calculated above.

Balancing forces horizontally at each joint shows that the horizontal component of tension is the same in each section of the cable : $$T_1\cos\theta_1=T_2\cos\theta_2=T_3\cos\theta_3=...$$ This shows that the maximum tension occurs where $\cos\theta$ has the smallest value, which is where the angle is maximum - ie at the ends. The maximum tension is in the outermost cable : you already calculated it above.

**Calculation**

Balance forces vertically and horizontally in turn at the joints A, B, C, D shown in the following diagram, starting from the outer end D :

At the 4 end-points D of the two horizontal cables, the vertical component of the tension $T_1$ in the 1st section of the cable is balanced by a reaction force equal to $\frac14 W$ where $W=4.80\times 10^4 kgwt=4.70\times 10^5 N$ is the full weight of the bridge. $$T_1\sin45^{\circ}=\frac{1}{\sqrt2}T_1=\frac14 W$$ $$T_1=\frac{\sqrt2}{4}W=1.70\times 10^4 kgwt=1.66\times 10^5N$$

At joint B the balance of vertical and horizontal forces gives $$T_1\sin\theta_1=T_2\sin\theta_2+\frac{1}{12}W$$ $$T_1\cos\theta_1=T_2\cos\theta_2$$ from which we get $$\frac{T_2\sin\theta_2}{T_2\cos\theta_2}=\tan\theta_2=1-\frac{W}{12T_1\sin\theta_1}=1-\frac{W}{3W}=\frac23$$ $$\theta_2=0.588 rad=33.7^{\circ}$$ $$\cos\theta_2=\frac{3}{\sqrt{13}}, \sin\theta_2=\frac{2}{\sqrt{13}}$$ $$T_2=\frac{T_1\cos\theta_1}{\cos\theta_2}=\frac{\sqrt{13}}{12}W=5.10\times 10^4 kgwt=5.00\times 10^5 N$$

At joint A we have similar forces so we can immediately write down that $$\tan\theta_3=1-\frac{W}{12T_2\sin\theta_2}=1-\frac{W}{2W}=\frac12$$ $$\theta_3=0.464rad=26.6^{\circ}$$ $$\cos\theta_3=\frac{2}{\sqrt5}, \sin\theta_3=\frac{1}{\sqrt5}$$ $$T_3=\frac{T_1\cos\theta_1}{\cos\theta_3}=\frac{\frac14}{\frac{2}{\sqrt5}}W=\frac{\sqrt5}{8}W=1.34\times 10^4 kgwt=1.31\times 10^5 N$$

We do not need to go any further. I leave it to you to calculate the lengths of the vertical cables.

**Note :** There may be something wrong with my calculation because the largest decrease is from $\theta_3=26.6^{\circ}$ to $\theta_4=0^{\circ}$. The largest decrease should be at the outside and the smallest at the inside. I am too tired to check this right now.

edited Sep 27 by JD_PM