Determine whether the following $H_0$ can be accepted or not using hypothesis testing

$$H_0 : p=0.5$$

$$H_1 : p > 0.5$$

where p is the probability of heads from a coin flip.

Let $$W_1$$ be the number of heads from 10 coin flips and and $$W_2$$ be the number of heads from 1000 coin flips.

Accept $$H_0$$ if $$W_1 \leq c_1$$ and $$W_2 \leq c_2$$, otherwise accept $$H_1$$. Determine the value of $$c_1$$ and $$c_2$$, or in other words the number of heads, using a level 0.05 test so that we may accept $$H_0$$.

I know that coin flips follow a binomial distribution, however I am not sure how to determine the two values using that. So far I've only solved problems with normal distributed random events. I am aware of CLT but is n=10 large enough to use it? For n=1000 it seems more reasonable. So do I have to use the fact that they are binomial distributed or is there some formula or trick that I am unaware of? Any help is appreciated!

For $$n=1000$$ it is certainly reasonable to just use a normal approximation. Be careful to use a one-tailed test.
For $$n$$ too small to use CLT (10 is on the small side), you can go back to the basic definition of a binomial distribution to write $$\alpha=\Pr(W_1>c_1\,|\,H_0)=\sum_{i=c_1+1}^{10}\binom{10}{i}\left(\frac{1}{2}\right)^{10}\text{,}$$ and solve for the smallest value of $$c_1$$ that leaves this probability below 0.05.
Note that we usually do not talk about "accepting" $$H_0$$ so much as failing to reject it, because not (enough) evidence that $$H_0$$ is false is not the same as evidence that $$H_0$$ is in fact true, in this frequentist setting.