Let the two points be $P(5,2), \quad Q(2,a), \quad O(0,0).$ The line segment $PQ$ subtends an angle $\frac{\pi}{4}$ at the origin, so $\angle POQ=\frac{\pi}{4}.$

Use the formula for angle between vectors $\overrightarrow{OP}$ and $\overrightarrow{OQ}$: $\cos\theta=\frac{\overrightarrow{OP}\cdot \overrightarrow{OQ}}{|OP|\,|OQ|}.$ Here, $\overrightarrow{OP}=(5,2), \qquad \overrightarrow{OQ}=(2,a).$

Compute dot product and magnitudes: $\overrightarrow{OP}\cdot \overrightarrow{OQ}=5\cdot 2+2a=10+2a,$ $|OP|=\sqrt{5^2+2^2}=\sqrt{29},$ $|OQ|=\sqrt{2^2+a^2}=\sqrt{4+a^2}.$

Since $\theta=\frac{\pi}{4}$, we have $\cos\frac{\pi}{4}=\frac{1}{\sqrt{2}}.$ Therefore, $\frac{10+2a}{\sqrt{29}\sqrt{4+a^2}}=\frac{1}{\sqrt{2}}.$

Cross-multiply: $\sqrt{2}(10+2a)=\sqrt{29(4+a^2)}.$ Squaring both sides, $2(10+2a)^2=29(4+a^2).$

Expand: $2(100+40a+4a^2)=116+29a^2,$ $200+80a+8a^2=116+29a^2.$ Rearranging, $21a^2-80a-84=0.$

Solve the quadratic: $21a^2-80a-84=0.$ Product of roots is $\frac{-84}{21}=-4.$

The question asks for the absolute value of the product of all possible values of $a$: $|-4|=4.$

Hence the correct option is $\boxed{4}.$