How do I find the solution $u(x,t)$ to this problem?
$$\left\{ \begin{array}{l l} u_t - ku_{xx} = 0 & \quad 0<x<\pi, t>0, \\ \quad u(x,0) = 3\sin x+6\sin 5x &\quad 0\leq x \leq \pi, \\ \quad u(0,t) = u(\pi,t)=0 &\quad t \geq 0. \end{array} \right. $$
By linearity of this differential equation, it is enough to solve these two initial-value problems separately and then add the two solutions: $$ \left\{ \begin{array}{l l} u_t - ku_{xx} = 0 & \quad 0<x<\pi, t>0, \\ \quad u(x,0) = 3\sin x &\quad 0\leq x \leq \pi, \\ \quad u(0,t) = u(\pi,t)=0 &\quad t \geq 0. \end{array} \right. $$ $$ \left\{ \begin{array}{l l} u_t - ku_{xx} = 0 & \quad 0<x<\pi, t>0, \\ \quad u(x,0) = 6\sin 5x &\quad 0\leq x \leq \pi, \\ \quad u(0,t) = u(\pi,t)=0 &\quad t \geq 0. \end{array} \right. $$ The second one will admit a solution of the form $$ u(x,t) = e^{at} 6\sin(5x) \tag 1 $$ provided you get the right value of $a.$ So find $u_t$ and $u_{xx}$ with $u$ as in $(1)$ in order to figure out what number $a$ has to be. The first initial-value problem above is treated similarly, but the value of $a$ will be different.
That this works out so neatly is the reason why Fourier series are the key to understanding diffusion equations of this sort.