Solve for $u(.5,13)$: $$u_{tt}=4u_{xx}\quad 0<x<1$$ $$u(x,0)=x, \ u_t(x,0)=1$$ $$u(0,t)=0, \ u_x(1,t)=0. $$
Using D'Alembert formula I got that I must take the odd extension and then apply the even extension since we have mixed boundary conditions which will give me a function of period $4$. I got that $\frac12[\bar{f}(x+2t)+\bar{f}(x-2t)]=\bar{f}(2.5)$, but the answer is $-\frac12$. Why is that?
Your reasoning is correct, but apparently unfinished: it does lead to the answer $-1/2$.
Let $\bar f$ be the (extended) initial position and $\hat g$ the (extended) initial velocity. Then $$ u(0.5, 13)= \frac12(\bar f(0.5-26)+\bar f(0.5+26))+\frac{1}{4}\int_{0.5-26}^{0.5+26}\bar g(s)\,ds $$ Since both $\bar f$ and $\bar g$ are $4$-periodic, and the integral over $g$ over its period is zero, we can simplify the above as $$ u(0.5, 13)= \frac12(\bar f(0.5-2 )+\bar f(0.5+2 ))+0 = \bar f(0.5+2) $$ Also, $2$ is the anti-period of $f$, so $\bar f(2.5)=-f(0.5)=-0.5$.