Step 1: Understand the problem.
We must find $\alpha$ where $\sin^{-1}\left(\frac{3}{5}\right) + \cos^{-1}\left(\frac{12}{13}\right) = \sin^{-1}\alpha$. The idea is to name each inverse term as an angle, then add the angles using the sine addition formula.
Step 2: Name the angles.
Let $A = \sin^{-1}\left(\frac{3}{5}\right)$, so $\sin A = \frac{3}{5}$. Let $B = \cos^{-1}\left(\frac{12}{13}\right)$, so $\cos B = \frac{12}{13}$. Both $A$ and $B$ are acute, so all their ratios are positive.
Step 3: Find the missing ratios.
For angle $A$: using a 3-4-5 idea, $\cos A = \frac{4}{5}$. For angle $B$: using a 5-12-13 idea, $\sin B = \frac{5}{13}$.
Step 4: Use the sine addition formula.
The sum is $\sin^{-1}\alpha = A + B$, so $\alpha = \sin(A+B)$. The formula is \[ \sin(A+B) = \sin A\cos B + \cos A\sin B. \]
Step 5: Plug in the values.
\[ \sin(A+B) = \frac{3}{5}\cdot\frac{12}{13} + \frac{4}{5}\cdot\frac{5}{13} = \frac{36}{65} + \frac{20}{65}. \]
Step 6: Add and finish.
\[ \frac{36}{65} + \frac{20}{65} = \frac{56}{65}. \] So $\alpha = \frac{56}{65}$. \[ \boxed{\alpha = \dfrac{56}{65}} \]