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One thing to do is to try to express these in terms of squares. Note that $$12x(3x-1)=36x^2-12x=(6x-1)^2-1$$ so that your equations become $$a_1^2+b_1^2=c_1^2+1$$ and $$a_1^2-b_1^2=d_1^2-1$$ where $a_1=6a-1$ etc. Then the variables $a_1$ etc are constrained to be congruent to $5$ modulo $6$.
Homogenizing these gives $$X^2+Y^2=Z^2+T^2$$ and $$X^2-Y^2=Z^2-T^2.$$ Seraching Searching for rational solutions of your equation is essentially looking for rational points on the intersection of these two quadrics in $\mathbf{P}^3$. In general the intersection of two quadrics in $\mathbf{P}^3$ is an elliptic curve, so it looks like your problem will boil down to something like finding the integer points on an elliptic curve.
Added There's a blunder in the above: I must thank Fedor for noticing that the second equation should be $$X^2-Y^2=W^2-T^2.$$ So the variety is the intersection of two quadrics in $\mathbf{P}^4$. Hartshorne mentions in passing that in general this construction gives a del Pezzo surface. Del Pezzo surfaces are rational so there should be a birational parametrizion (in terms of two affine parameters) of the rational solutions to the original pair of equations.
One thing to do is to try to express these in terms of squares. Note that $$12x(3x-1)=36x^2-12x=(6x-1)^2-1$$ so that your equations become $$a_1^2+b_1^2=c_1^2+1$$ and $$a_1^2-b_1^2=d_1^2-1$$ where $a_1=6a-1$ etc. Then the variables $a_1$ etc are constrained to be congruent to $5$ modulo $6$.
Homogenizing these gives $$X^2+Y^2=Z^2+T^2$$ and $$X^2-Y^2=Z^2-T^2.$$ Seraching for rational solutions of your equation is essentially looking for rational points on the intersection of these two quadrics in $\mathbf{P}^3$. In general the intersection of two quadrics in $\mathbf{P}^3$ is an elliptic curve, so it looks like your problem will boil down to something like finding the integer points on an elliptic curve.