1. Let $\alpha=\frac1p,\quad \beta=\frac1q,\quad \gamma=\frac1r,\quad \delta=\frac1s$ and it is given that $\alpha,\beta,\gamma,\delta$ are in A.P.

2. First equation: $x^2-8ax+2a=0$ has roots $p,r$.

   By Vieta, $p+r=8a,\qquad pr=2a$

   Therefore, $\frac1p+\frac1r=\frac{p+r}{pr}=\frac{8a}{2a}=4$ So, $\alpha+\gamma=4$

3. Second equation: $x^2+12bx+6b=0$ has roots $q,s$.

   By Vieta, $q+s=-12b,\qquad qs=6b$

   Therefore, $\frac1q+\frac1s=\frac{q+s}{qs}=\frac{-12b}{6b}=-2$ So, $\beta+\delta=-2$

4. Since $\alpha,\beta,\gamma,\delta$ are in A.P., let their common difference be $d$. Then $\beta=\alpha+d,\quad \gamma=\alpha+2d,\quad \delta=\alpha+3d$

   Using the sums found above: $\alpha+\gamma=\alpha+(\alpha+2d)=2\alpha+2d=4$ $\beta+\delta=(\alpha+d)+(\alpha+3d)=2\alpha+4d=-2$

   Subtracting, $2d=-6\implies d=-3$

   Then from $2\alpha+2d=4$ we get $2\alpha-6=4\implies 2\alpha=10\implies \alpha=5$

   Hence the A.P. is $5,2,-1,-4$

   So, $\left\{\frac1p,\frac1r\right\}=\{5,-1\},\qquad \left\{\frac1q,\frac1s\right\}=\{2,-4\}$

5. Now use products.

   For $p,r$: $\frac1p\cdot\frac1r=\frac1{pr}=\frac1{2a}$ But from above, $5\cdot(-1)=-5$ Hence, $\frac1{2a}=-5\implies 2a=-\frac15\implies a=-\frac1{10}$ Therefore, $a^{-1}=-10$

6. For $q,s$: $\frac1q\cdot\frac1s=\frac1{qs}=\frac1{6b}$ But from above, $2\cdot(-4)=-8$ Hence, $\frac1{6b}=-8\implies 6b=-\frac18\implies b=-\frac1{48}$ Therefore, $b^{-1}=-48$

7. Therefore, $a^{-1}-b^{-1}=(-10)-(-48)=38$

So the required integer is $\boxed{38}$