Don't think about $a+b+c+d$ as a known equation!

$\begin{cases} a+b+c-d = 1 \space ...(1) \\ a+b-c+d =2 \space ...(2) \\ a-b+c+d = 3 \space ...(3) \\ -a+b+c+d = 4 \space ...(4) \end{cases}$

$(1) + (2) + (3) + (4) \Rightarrow \space 2(a+b+c+d) = 10 \Rightarrow \space a+b+c+d = 5 \space ...(5)$

$(5) - (1)$ gives $d= 2$ .

$(5) - (2)$ gives $c = \frac{3}{2}$ .

$(5) - (3)$ gives $b = 1$ .

$(5) - (4)$ gives $a = \frac{1}{2}$ .

Then, we have $a² = \frac{1}{4}; \space b² = 1; \space c² = \frac{9}{4}; \space d² = 4$ .

Hence, we have $8a² + 9b² + 12c² + 13d² = 2 + 9 + 27 + 52 = \boxed{90}$ .