Say we need a value $s$ such that: $$P(Z\le s) = 0.995$$ Using the cumulative $Z$ table (with 3 decimal-place accuracy), I realized you can get several estimated values of $s$:
These include $$2.574, 2.575, 2.576, 2.577, 2.578, 2.579$$ Due to the limit of accuracy, all these values are correct.
My question is, is there any better way of getting a more "accurate" value or are all of these values acceptable (at the A-Level/high school level)?
It depends both on the number of digits in the $Z$ and $\Phi(Z)$ values in your table. The one in Wikipedia only shows $Z$ to two places past the decimal but shows the $\Phi(Z)$ values to five places. Looking at that table, any $Z$ between $2.55$ and $2.61$ returns a value which would round the probability to $0.995$ to three places. The one that is closest is $2.58$. You can interpolate the table to get more accuracy. This table gives $P(2.57)=0.99492, P(2.58)=0.99506$. The simplest is linear interpolation, which essentially takes these two points as exact and fits a line through them. We find $z(0.995)=2.57+\frac {0.99500-0.99492}{0.99506-0.99492}(2.58-2.57)\approx 2.575$. By looking at successive values in the area you can see that the function is quite linear in this region. When I check with Alpha I find the correct value is about $2.57583$. If the table you have only gives $\Phi(Z)$ to three places there is not much you can do besides take the middle one.