1. We need to evaluate $I=\int_{-1}^{1}\frac{\cos(\alpha x)}{1+3^x}\,dx$ and use the condition $I=\frac{2}{\pi}$.

2. Use the standard symmetry trick: let $f(x)=\frac{\cos(\alpha x)}{1+3^x}.$ Then $I=\int_{-1}^{1} f(x)\,dx = \int_{-1}^{1} \frac{f(x)+f(-x)}{2}\,dx.$ So first compute $f(-x)$: f(-x)=\frac{\cos(-\alpha x)}{1+3^{-x}}=rac{\cos(\alpha x)}{1+3^{-x}}. Since $\frac{1}{1+3^{-x}}=\frac{3^x}{1+3^x},$ we get $f(-x)=\cos(\alpha x)\cdot \frac{3^x}{1+3^x}.$

Thus, $f(x)+f(-x)=\cos(\alpha x)\left(\frac{1}{1+3^x}+\frac{3^x}{1+3^x}\right)=\cos(\alpha x).$ Hence, $I=\frac12\int_{-1}^{1}\cos(\alpha x)\,dx.$

3. Now evaluate: \int_{-1}^{1}\cos(\alpha x)\,dx=\left[\frac{\sin(\alpha x)}{\alpha}\right]_{-1}^{1}=rac{\sin\alpha-\sin(-\alpha)}{\alpha}=\frac{2\sin\alpha}{\alpha}. Therefore, $I=\frac12\cdot \frac{2\sin\alpha}{\alpha}=\frac{\sin\alpha}{\alpha}.$

4. Given that $\frac{\sin\alpha}{\alpha}=\frac{2}{\pi}.$ Now test the options.

• A: $\alpha=\frac{\pi}{2}$ $\frac{\sin(\pi/2)}{\pi/2}=\frac{1}{\pi/2}=\frac{2}{\pi}$ This works.

• B: $\alpha=\frac{\pi}{4}$ $\frac{\sin(\pi/4)}{\pi/4}=\frac{\frac{\sqrt2}{2}}{\pi/4}=\frac{2\sqrt2}{\pi}\neq \frac{2}{\pi}$ Not correct.

• C: $\alpha=\frac{\pi}{3}$ $\frac{\sin(\pi/3)}{\pi/3}=\frac{\frac{\sqrt3}{2}}{\pi/3}=\frac{3\sqrt3}{2\pi}\neq \frac{2}{\pi}$ Not correct.

• D: $\alpha=\frac{\pi}{6}$ $\frac{\sin(\pi/6)}{\pi/6}=\frac{\frac12}{\pi/6}=\frac{3}{\pi}\neq \frac{2}{\pi}$ Not correct.

5. Therefore, the correct option is $\boxed{\alpha=\frac{\pi}{2}}.$