The book I'm reading defines an event as a subset of the sample space, so if coin $1$ is an event then it must be a subset of the sample space. This subset must be all combinations of coin tosses given coin $1$, which is the same as the amount of coin tosses using coin $2$. What exactly is in the events (subsets of the sample space) coin $1$ and coin $2$ that makes them a partition is what I'm confused about. coin $1 = {x,y,z,...}$ and coin $2={w,r,t,...}$ such that coin $1 ∩ $ coin $2=∅$ and their union is the entire space of coin tosses.
An appropriate sample space to describe this system is $\{\tfrac 1 2,\tfrac 1 3\}{\times}\{{\rm T,H}\}^5$, that is the Cartesian product of outcomes for selecting a coin (denoted by its bias), and tossing that coin five times. Your random variable is the count of heads obtained. Let us call it $X$.
$X=0$ denotes the event of $\rm \{(\tfrac 12, T,T,T,T,T),(\tfrac 13, T, T, T, T, T)\}$ . That is the outcome of selecting the unbiased coin and then tossing five tails, or of selecting the biased coin and then tossing five tails. And so forth. The probability of this event is calculated as: $$\mathsf P(X=0)~=~ \tfrac 12(\tfrac 12)^5+\tfrac 12(\tfrac 23)^5$$
As you can see, we can partition the space as $\{\tfrac 12\}{\times}\{{\rm T,H}\}^5~\cup~\{\tfrac 13\}{\times}\{{\rm T,H}\}^5$, that is on the event of selecting which coin. Let $Y$ be the bias of the coin.
$Y=\tfrac 12$ describes the event $\rm \{(\tfrac 12, T, T, T, T, T), (\tfrac 12, T, T, T, T, H), \ldots, (\tfrac 12, H, H, H, H, H)\}$, etcetera. Thirty two outcomes of various probabilities. $\mathsf P(Y=\tfrac 12)=\tfrac 12=\mathsf P(Y=\tfrac 13)$ because selecting the coin itself is made without bias.
That's the meat. The crunch of it is that:
$$\begin{align}\mathsf P(X\geq 3) ~=~& \mathsf P(Y=\tfrac 12\cap X\geq 3)+\mathsf P(Y=\tfrac 13\cap X\geq 3)
\\[1ex] =~& \tfrac 12 \Big(\sum_{x=3}^5 \binom 5 x (\tfrac 12)^x(\tfrac 12)^{5-x}\Big)+\tfrac 12\Big(\sum_{x=3}^5 \binom 5 x (\tfrac 13)^x(\tfrac 23)^{5-x}\Big)
\\[1ex] =~& \tfrac 14+\tfrac 1 2 \Big(\binom 5 3\tfrac{2^2}{3^5}+ \binom 5 4\tfrac{2}{3^5}+\binom 5 5\tfrac 1{3^5} \Big)
\\[1ex] =~& \tfrac 14+\tfrac{40+10+1}{2\cdot 3^5}
\\[1ex] =~& \tfrac {115}{324}
\\[1ex] =~& 0.35\dot{\overline{493827160}}
\end{align}$$