What is $P(x \ge 10)$ given $n = 40$ and $p = 0.2$?
I calculated that the mean $= 8$ and sd $= 2.53$ Therefore $z = 0.791$, so $P = 1 - 0.7852 = 0.2148$.
How ever the answer shows as $0.2682$?
I see that there is a condition that $np$ and $np(1 - p)$ must be at least $10$ for the mean what is the alternative/ is this relevant?
$$ \Pr(X\ge 10) = \Pr(X>9). $$ The usual continuity correction approximated this by $\Pr(X\ge 9.5)$ and treating $X$ as normally distributed with the same expected value and the same standard deviation as the binomially distributed random variable. \begin{align} \Pr\left(\frac{X-8}{\sqrt{6.4}} \ge \frac{9.5-8}{\sqrt{6.4}}\right) \approx \Pr(Z\ge 0.5929271) = \Phi(0.5929271) \approx 0.276615. \end{align} My software is telling me that for the binomial distribution, rather than the normal approximation, one gets $\Pr(X\ge 10) \approx 0.2682229$.
Your biggest problem is not using the continuity correction. But also, you should never do more rounding than is absolutely necessary until the very last step, unless you know what the effect of rounding at an early stage will be on the bottom line. You did that twice.