The correct answer is $\boxed{\frac{1}{24}\left[\begin{array}{cc}4-3i&i\-i&4+3i\end{array}\right]}$.
To find the inverse of a 2×2 matrix, we can use the formula $A^{-1}=\frac{1}{|A|}\left[\begin{array}{cc}a_{2,2}&-a_{1,2}\\a_{2,1}&a_{1,1}\end{array}\right]$, where $|A|$ is the determinant of $A$.
In this case, $A=\left[\begin{array}{cc}4+3i&-i\\i&4-3i\end{array}\right]$. The determinant of $A$ can be calculated as follows:
$$|A|=4^2-(3i)^2-(-i)(4-3i)=4^2+3^2+4i=25$$
Therefore, the inverse of $A$ is $\frac{1}{|A|}\left[\begin{array}{cc}4-3i&-i\\i&4+3i\end{array}\right]=\frac{1}{25}\left[\begin{array}{cc}4-3i&-i\\i&4+3i\end{array}\right]$.