1. Let the G.P. be $a_1=a,\, a_2=ar,\, a_3=ar^2,\, a_4=ar^3,\, a_5=ar^4,\, a_6=ar^5$ where, since the terms are increasing and positive, we have $a>0,\, r>1.$

2. Use the condition $a_1a_5=28$: $a\cdot ar^4=28$ $a^2r^4=28$ $\left(ar^2\right)^2=28$ Hence, $ar^2=\sqrt{28}=2\sqrt{7}$ since all terms are positive.

3. Use the condition $a_2+a_4=29$: $ar+ar^3=29$ $ar(1+r^2)=29$

4. Write this in terms of $ar^2$: $ar+ar^3=ar\left(1+r^2\right)=\frac{ar^2}{r}(1+r^2)=ar^2\left(r+\frac{1}{r}\right)$ So, $ar^2\left(r+\frac{1}{r}\right)=29$ $2\sqrt{7}\left(r+\frac{1}{r}\right)=29$ $r+\frac{1}{r}=\frac{29}{2\sqrt{7}}$

5. Observe that $\frac{29}{2\sqrt{7}}=\frac{28+1}{2\sqrt{7}}=\frac{\sqrt{7}}{2}+\frac{1}{2\sqrt{7}}=\sqrt{7}+\frac{1}{\sqrt{7}}$ Hence, $r+\frac{1}{r}=\sqrt{7}+\frac{1}{\sqrt{7}}$ Since $r>1$, we get $r=\sqrt{7}.$

6. Now find $a$ using $ar^2=2\sqrt{7}$: $a\cdot 7=2\sqrt{7}$ $a=\frac{2\sqrt{7}}{7}.$

7. Compute $a_6=ar^5$: $a_6=a(\sqrt{7})^5=a\cdot 49\sqrt{7}$ $a_6=\frac{2\sqrt{7}}{7}\cdot 49\sqrt{7}$ $a_6=2\cdot 7\cdot 7=98.$

8. Check quickly: $a_1a_5=a\cdot ar^4=(ar^2)^2=(2\sqrt{7})^2=28,$ $a_2+a_4=ar+ar^3=ar(1+r^2)=\frac{2\sqrt{7}}{7}\cdot \sqrt{7}\cdot 8=2\cdot 8+?$ More directly, $a_2=ar=2,\quad a_4=ar^3=14,$ so actually with $r=\sqrt7$ and $a=\frac{2\sqrt7}{7}$, $a_2=\frac{2\sqrt7}{7}\cdot \sqrt7=2,$ $a_4=\frac{2\sqrt7}{7}\cdot 7\sqrt7=14,$ giving $a_2+a_4=16,$ which does not satisfy the given condition $29$.

   So the identification in Step 5 was incorrect. Let us solve properly.

9. From $ar^2=2\sqrt7$ and $ar+ar^3=29,$ divide the second by the first: $\frac{ar+ar^3}{ar^2}=\frac{29}{2\sqrt7}$ $\frac{1}{r}+r=\frac{29}{2\sqrt7}.$ So $r$ satisfies $r+\frac1r=\frac{29}{2\sqrt7}.$ Multiply by $r$: $r^2-\frac{29}{2\sqrt7}r+1=0.$

10. We need $a_6=ar^5=(ar^2)r^3=2\sqrt7\,r^3$.

    Instead of solving for irrational $r$ directly, note $\left(r+\frac1r\right)^2=r^2+2+\frac1{r^2}.$ Let $s=r+\frac1r=\frac{29}{2\sqrt7}.$ Then r^2+\frac1{r^2}=s^2-2=rac{841}{28}-2=\frac{785}{28}.

11. Since $r>1$, take the larger root of $r^2-sr+1=0.$ But checking the options, it is clear $a_6$ should likely be much larger than 98 only if there is a typo in interpretation. Let us derive directly using another substitution.

12. Let $x=a_2=ar,\quad y=a_4=ar^3.$ Then because it is a G.P., $a_3^2=a_2a_4=xy.$ Also, $a_1a_5=a_3^2=28.$ Hence, $xy=28.$ And given, $x+y=29.$

13. So $x,y$ are roots of $t^2-29t+28=0.$ Factorizing, $t^2-29t+28=(t-1)(t-28)=0.$ Thus, $\{x,y\}=\{1,28\}.$

14. Since the G.P. has increasing positive terms, we must have $a_2<a_4,$ so $a_2=1,\quad a_4=28.$

15. Then the common ratio satisfies $\frac{a_4}{a_2}=r^2=\frac{28}{1}=28.$ Thus, $r=\sqrt{28}=2\sqrt7$ (positive and $>1$).

16. Now $a_6=a_4\cdot r^2=28\cdot 28=784.$

Therefore, the correct answer is $\boxed{784}$ which corresponds to Option B.