Here is my solution, which I am hoping can either be verified or corrected.
$B(t)$ is normal with mean 0 and variance $t$.
$P(B(4)>1\;|\;B(8)=0) = P(B(4)+B(8)>1+0\;|\;B(8)=0) = P(B(12)>1) = P(Z>\frac{1}{\sqrt{12}}) = 1-P(Z<0.29) = 1-0.6141 = 0.3859$.
Because the pair $(B(4),B(8))$ has a joint normal distribution, the (regular) conditional distribution of $B(4)$, given that $B(8) = z$ is normal with mean $z/2$ and variance $2$.
Alternatively, again because $(B(4),B(8))$ has a joint normal distribution, the fact that $B(4)-.5B(8)$ and $B(8)$ are uncorrelated implies that they are independent, and the former has a normal distribution with mean $0$ and variance $2$. Thus, $B(4) = [B(4)-.5B(8)]+.5B(8)$, conditional on $B(8) = z$ is normal with mean $z/2$ and variance that of $B(4)-.5B(8)$, namely $2$.