the product of the three first terms equals the 2nd term cubed. thus the 2nd term is 4. Let the 3 terms now be 4/a, 4, and 4a. We have a+1/a=5/2, which has only solutions 2 and 1/2. Thus the two solutions are 2, 4, 8, and 8, 4, 2.

Let $a,b,c$(in this order) be the first three terms. We know that $abc=64$ and $a+b+c=14$ Nice they're in a geometric progression we have that $b=ak$ and $c=ak^2$ where $k$ is our constant so: $abc=64$ $\implies$ $a^3k^3=64$ $\implies$ $ak=4$ and $a+ak+ak^2=14$ which is the same as $a+4k=10$ so $a=10-4k$ and using this with $ak=4$ you'll get $2k^2-5k+2=0$ and after some trivial algebra you get the following solutions: $k=2$ then $(a,b,c)=(2,4,8)$ $k=\frac{1}{2}$ then $(a,b,c)=(8,4,2)$

Let the terms be $a/r, a, ar$. Then $a^3 = 64 \implies a = 4$, and $r + 1/r = 5/2 \implies r = 2, 1/2$, yielding $(2, 4, 8) and (8, 4, 2)$.