Why is the Shannon entropy 0.94 in this example?
Back to the definitions:
$$H(L\mid A) = \sum_a p(A=a) H(L \mid A=a).$$

As you compute, $P(A=true)=6/8$ and $P(A=false)=2/8$.

However, you don't compute $H(L\mid A=true)$ but instead compute $P(L=positive\mid A=true)$.  [and the same for $A=false$.]. 

With standard definition of $H()$ we get,  

$$H(L\mid A=true) = - 4/6\log_2(4/6) - 2/6\log_2(2/6) = 0.9182958$$
$$H(L\mid A=false) = H(1/2) = 1$$

And thus, $H(L\mid A) = 6/8 \times 0.918 + 2/8\times 1 = 0.938 \approx 0.94$.