This problem follows similar logic to the first problem.

(a) We let $\lambda = \omega^2$. We can then proceed as in the first problem (check that post if you would like to see the steps) to get that: \begin{equation}

X(x) = A\cos(\omega{}x) + B\sin(\omega{}x) \end{equation}

With the derivative being: \begin{equation}

X'(x) = -A\omega{}\sin(\omega{}x) + B\omega{}\cos(\omega{}x)

\end{equation}

This time we have a Dirichlet condition that $X(0) = 0$. So plugging this in: \begin{equation}

X(0) = A = 0 \end{equation}

Therefore $A=0$. Thus the solution is: \begin{equation}

X_n = \sin(\omega{}_nx) \end{equation}

We have chosen $B=1$ for convenience.

Edit: the similar tangent relation is, as you can easily show using the boundary condition on the right end, is: \begin{equation}

\tan(\omega{}l) = -\frac{\omega{}}{\beta{}} \end{equation}