= Solution
For independent small uncertainties interpreted as <standard deviations>, <propagation of uncertainty> gives
$$
\sigma_N^2=(\sin\alpha)^2\sigma_n^2+(n\cos\alpha)^2\sigma_\alpha^2,\qquad \left(\frac{\sigma_N}{N}\right)^2=\left(\frac{\sigma_n}{n}\right)^2+\cot^2\alpha\,\sigma_\alpha^2.
$$
The angular uncertainty is in radians. Since $\sigma_n/n=0.075/1.35=1/18$, the allowed angular contribution to the fractional uncertainty of the <numerical aperture> is only $\sqrt{0.06^2-(1/18)^2}$. Therefore
$$
\boxed{\sigma_\alpha\leq\tan60^\circ\sqrt{0.06^2-(1/18)^2}=0.0393\,\mathrm{rad}\simeq2.25^\circ.}
$$
The uncertainty in the <refractive index> already contributes $5.56\%$. If the quoted errors instead mean strict worst-case bounds, first-order absolute errors must be added rather than added in quadrature; that convention would require $|\delta\alpha|\lesssim\tan60^\circ(0.06-1/18)=0.441^\circ$. The $2.25^\circ$ result uses the usual independent statistical-error convention.
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