1. Use the tangent condition for the hyperbola

Given hyperbola: $\frac{x^2}{a^2}-\frac{y^2}{16}=1$

Given line: $2x-y+1=0 \quad \Rightarrow \quad y=2x+1$

For this line to be tangent to the hyperbola, substituting $y=2x+1$ into the hyperbola must give a quadratic in $x$ with discriminant $0$.

So, $\frac{x^2}{a^2}-\frac{(2x+1)^2}{16}=1$

$\frac{x^2}{a^2}-\frac{4x^2+4x+1}{16}=1$

Multiply by $16a^2$: $16x^2-a^2(4x^2+4x+1)=16a^2$

$16x^2-4a^2x^2-4a^2x-a^2-16a^2=0$

$4(4-a^2)x^2-4a^2x-17a^2=0$

For tangency, $\Delta=0$

$(-4a^2)^2-4\cdot 4(4-a^2)(-17a^2)=0$

$16a^4+272a^2(4-a^2)=0$

$16a^2\big(a^2+17(4-a^2)\big)=0$

$16a^2(68-16a^2)=0$

Since $a\neq 0$, $68-16a^2=0$

$a^2=\frac{17}{4}$

Hence, $a=\frac{\sqrt{17}}{2}$

2. Check each option for being sides of a right triangle

A set of three positive numbers can be sides of a right triangle if the square of the largest equals the sum of squares of the other two.

Option A: $a,4,1$

Here, $a^2=\frac{17}{4}=4.25$

Largest side is $4$. Check: $4^2 \stackrel{?}{=} a^2+1^2$ $16 \stackrel{?}{=} \frac{17}{4}+1=\frac{21}{4}$ This is false.

So A cannot be sides of a right triangle.

Option B: $2a,4,1$

$2a=\sqrt{17}$ So squares are: $(2a)^2=17,\quad 4^2=16,\quad 1^2=1$

Now, $17=16+1$ Hence these can be sides of a right triangle.

So B can be sides of a right triangle.

Option C: $a,4,2$

Largest side is $4$. Check: $4^2 \stackrel{?}{=} a^2+2^2$ $16 \stackrel{?}{=} \frac{17}{4}+4=\frac{33}{4}$ False.

Also other possibilities clearly fail. So C cannot be sides of a right triangle.

Option D: $2a,8,1$

Squares are: $(2a)^2=17,\quad 8^2=64,\quad 1^2=1$

Check largest side $8$: $64 \stackrel{?}{=} 17+1=18$ False.

So D cannot be sides of a right triangle.

3. Final conclusion

The sets that cannot be sides of a right angled triangle are: $\boxed{A,\ C,\ D}$

This matches the stored correct answer.