Half-life and shelf-life of second-order reaction

Your mathematics is correct until you calculated for specific cases. Let's go back to your $n$th order version:

$$\frac{[\ce{A}]^{1-n}}{1-n} = -k_nt + \frac{[\ce{A}]_\circ^{1-n}}{1-n} \ \text{where } n \ne 1 \tag1$$

At this point, since you are working on second order kinetic, it is easy if you substitute $n = 2$ in equation $(1)$:

$$\frac{[\ce{A}]^{-1}}{-1} = -k_2t + \frac{[\ce{A}]_\circ^{-1}}{-1} $$

Once simplify, it becomes:

$$\frac{1}{[\ce{A}]} = k_2t + \frac{1}{[\ce{A}]_\circ} \tag2$$

For self life, substitute $t = t_{10\%}$ and ${[\ce{A}]} = 0.9{[\ce{A}]_\circ}$ in equation $(2)$:

$$\frac{1}{0.9[\ce{A}]_\circ} = k_2t_{10\%} + \frac{1}{[\ce{A}]_\circ} $$

$$\therefore \; t_{10\%} = \frac{1}{k_2}\left(\frac{1}{0.9[\ce{A}]_\circ} - \frac{1}{[\ce{A}]_\circ} \right) = \frac{1}{k_2[\ce{A}]_\circ}\left(\frac{1}{0.9} - 1 \right) = \frac{1}{k_2[\ce{A}]_\circ}\left(\frac{1}{9} \right) = \bbox[yellow]{\frac{0.11}{k_2[\ce{A}]_\circ}}$$

Similarly, for half life, substitute $t = t_{50\%}$ and ${[\ce{A}]} = 0.5{[\ce{A}]_\circ}$ in equation $(2)$:

$$\frac{1}{0.5[\ce{A}]_\circ} = k_2t_{50\%} + \frac{1}{[\ce{A}]_\circ} $$

$$\therefore \; t_{50\%} = \frac{1}{k_2}\left(\frac{1}{0.5[\ce{A}]_\circ} - \frac{1}{[\ce{A}]_\circ} \right) = \frac{1}{k_2[\ce{A}]_\circ}\left(\frac{1}{0.5} - 1 \right) = \frac{1}{k_2[\ce{A}]_\circ}\left(\frac{0.5}{0.5} \right) = \bbox[yellow]{\frac{1}{k_2[\ce{A}]_\circ}}$$