1. Interpret the statement

Let the two distinct positive numbers be $a$ and $b$.

• If $A_1, A_2$ are the two arithmetic means between $a$ and $b$, then $a, A_1, A_2, b$ are in A.P.

• If $G_1, G_2, G_3$ are the three geometric means between $a$ and $b$, then $a, G_1, G_2, G_3, b$ are in G.P.

We must find $G_1^4+G_2^4+G_3^4+G_1^2G_3^2.$

2. Find $A_1+A_2$

Since $a, A_1, A_2, b$ are in A.P., the common difference is $d=\frac{b-a}{3}.$ So, $A_1=a+d=a+\frac{b-a}{3}=\frac{2a+b}{3},$ $A_2=a+2d=a+\frac{2(b-a)}{3}=\frac{a+2b}{3}.$

Therefore, $A_1+A_2=\frac{2a+b}{3}+\frac{a+2b}{3}=a+b.$

3. Express the geometric means

Let the common ratio of the G.P. be $r$. Then $a, ar, ar^2, ar^3, ar^4=b.$ So, $G_1=ar,\quad G_2=ar^2,\quad G_3=ar^3,$ and $r^4=\frac{b}{a}.$

Now compute each term: $G_1^4=(ar)^4=a^4r^4=a^3b,$ $G_2^4=(ar^2)^4=a^4r^8=a^2b^2,$ $G_3^4=(ar^3)^4=a^4r^{12}=ab^3,$ because $r^8=(r^4)^2=(b/a)^2$ and $r^{12}=(r^4)^3=(b/a)^3$.

Also, $G_1^2G_3^2=(ar)^2(ar^3)^2=a^4r^8=a^2b^2.$

Hence, \begin{align*} G_1^4+G_2^4+G_3^4+G_1^2G_3^2 &= a^3b+a^2b^2+ab^3+a^2b^2 \ &= a^3b+2a^2b^2+ab^3 \ &= ab(a^2+2ab+b^2) \ &= ab(a+b)^2. \end{align*}

4. Relate with the options

Now, $G_1G_3=(ar)(ar^3)=a^2r^4=ab.$ Also, $A_1+A_2=a+b.$

Therefore, $\left(A_1+A_2\right)^2G_1G_3=(a+b)^2\cdot ab=ab(a+b)^2,$ which matches the required expression.

So the correct option is $\boxed{A}.$

5. Verification with stored answer

Stored correct answer: $A$

Derived answer: $A$

They agree.