1. Write the G.P. in standard form

Let the G.P. be $a_n = ar^{n-1}$ with $a>0$ and, since the terms are increasing positive numbers, we have $r>1.$

2. Use the condition on the 6th and 8th terms

We have $a_6+a_8=ar^5+ar^7=ar^5(1+r^2)=2. \qquad (1)$

3. Use the condition on the 3rd and 5th terms

Also, $a_3a_5=(ar^2)(ar^4)=a^2r^6=\frac19. \qquad (2)$

From (2), $ar^3=\frac13$ because $a>0, r>0$.

4. Substitute into equation (1)

Since $ar^5=(ar^3)r^2=\frac13 r^2,$ we get from (1): $\frac13 r^2(1+r^2)=2.$ So, $r^2(1+r^2)=6.$ Let $x=r^2$. Then $x+x^2=6$ $x^2+x-6=0$ $(x-2)(x+3)=0.$ Since $x=r^2>0$, we get $r^2=2.$ Hence $r=\sqrt2.$

Also, from $ar^3=\frac13$, $a=\frac{1}{3r^3}.$

5. Compute $a_2+a_4$ and $a_4+a_6$

We have $a_2=ar, \quad a_4=ar^3, \quad a_6=ar^5.$ Thus, $(a_2+a_4)=ar+ar^3=ar(1+r^2),$ $(a_4+a_6)=ar^3+ar^5=ar^3(1+r^2).$ Therefore, $(a_2+a_4)(a_4+a_6)=a^2r^4(1+r^2)^2.$

So, $6(a_2+a_4)(a_4+a_6)=6a^2r^4(1+r^2)^2. \qquad (3)$

6. Use $a^2r^6=\frac19$

From (2), $a^2r^6=\frac19 \implies a^2r^4=\frac{1}{9r^2}.$ Substitute into (3):