Let the geometric progression be $a,\, ar,\, ar^2,\, ar^3,\, ar^4,\, ar^5,\dots$ Since it is an increasing geometric series, we take $a>0,\quad r>1.$

Use the given product condition The 3rd and 5th terms are: $T_3=ar^2,\qquad T_5=ar^4$ Their product is 25: $T_3T_5=(ar^2)(ar^4)=a^2r^6=25$ $\Rightarrow (ar^3)^2=25$ Since terms are positive in an increasing GP, $ar^3=5.$

Use the given sum condition The 2nd and 6th terms are: $T_2=ar,\qquad T_6=ar^5$ Given: $ar+ar^5=\frac{25}{2}$ Factor out $ar^3$: $ar(1+r^4)=ar^3\left(r^{-2}+r^2\right)$ Using $ar^3=5$, $5\left(r^2+\frac{1}{r^2}\right)=\frac{25}{2}$ $\Rightarrow r^2+\frac{1}{r^2}=\frac{5}{2}$

Find $r+\frac1r$ We know $\left(r+\frac1r\right)^2=r^2+\frac1{r^2}+2=\frac{5}{2}+2=\frac{9}{2}$ So, $r+\frac1r=\frac{3}{\sqrt2}$ but this is not especially convenient directly. Instead, let $x=r^2.$ Then $x+\frac1x=\frac52$ $\Rightarrow 2x^2-5x+2=0$ $\Rightarrow (2x-1)(x-2)=0$ Hence, $x=2\quad \text{or} \quad x=\frac12.$ Since $r>1$, we must have $r^2=2 \Rightarrow r=\sqrt2.$

Find $a$ From $ar^3=5,$ and $r=\sqrt2$, $a(\sqrt2)^3=5$ $a(2\sqrt2)=5$ $a=\frac{5}{2\sqrt2}.$

Compute the required sum The 4th, 6th, and 8th terms are: $T_4=ar^3,\qquad T_6=ar^5,\qquad T_8=ar^7$ Their sum is $ar^3+ar^5+ar^7=ar^3(1+r^2+r^4).$ Using $ar^3=5$ and $r^2=2$, $=5(1+2+4)=5\cdot 7=35.$

Conclusion $\boxed{35}$ So the correct option is D.