a) Following the hint, I made a change of variables $r=x-\frac{l}{2}$. The beam equation and its boundary conditions then become:

$$u_{tt}+Ku_{rrrr}=0,~-\frac{l}{2}<r<\frac{l}{2}\\u(-\frac{l}{2},t)=u_{r}(-\frac{l}{2},t)= u(\frac{l}{2},t)=u_{r}(\frac{l}{2},t)=0$$

Letting $u=T(t)R(r)$, we see that $$T''+K\lambda T=0\\ R''''-\lambda R=0$$

Here we are assuming that $\lambda>0$, so let $\lambda=\omega^{4}$. Solving the corresponding characteristic equations and ODEs we get

$$T=A\cos(\sqrt{K}\omega^{2}t)+B\sin(\sqrt{K}\omega^{2}t)\\ R=Ce^{\omega r}+De^{-\omega r}+E\cos(\omega r)+F\sin(\omega r)$$

Using the 4 boundary conditions, the fact that $T$ is a constant with respect to $r$ and is not identically $0$ we arrive at the system of 4 equations below

$$Ce^{-\omega\frac{l}{2}}+De^{\omega\frac{l}{2}}+E\cos(\omega\frac{l}{2})-F\sin(\omega\frac{l}{2})=0\\

Ce^{-\omega\frac{l}{2}}-De^{\omega\frac{l}{2}}+E\sin(\omega\frac{l}{2})+F\cos(\omega\frac{l}{2})=0\\

Ce^{\omega\frac{l}{2}}+De^{-\omega\frac{l}{2}}+E\cos(\omega\frac{l}{2})+F\sin(\omega\frac{l}{2})=0\\

Ce^{\omega\frac{l}{2}}-De^{-\omega\frac{l}{2}}-E\sin(\omega\frac{l}{2})+F\cos(\omega\frac{l}{2})=0$$

We can convert this into a matrix, and it only has non-trivial solutions when the determinant of the matrix is $0$. This computation is tedious, but the final condition that $\omega$ must satisfy is

$$\cos(\omega l)\cosh(\omega l)=1$$