For any future visitors, the expression given for the probability of at least $m$ out of $n$ events $(A_1, \ldots ,A_n)$ in An Introduction to Probability by William Feller is this:
$$P_m = S_m - \binom{m}{1}S_{m+1}+\binom{m+1}{2}S_{m+2}-\ldots\pm\binom{n-1}{m-1}S_n$$
where $$\displaystyle S_k = \sum_{1\leq i_1< i_2\ldots< i_k\leq n}P(A_{i_1}\cap A_{i_2}\cap\ldots \cap A_{i_k})$$
It's certainly not pretty, but it is general.
It is proved by finding the expression for exactly $m$ events and then adding the expressions from $m$ to all $n$ using induction.
Following your and Stefanos' comments and your edit, I'm revising my answer. I think the first step is to precisely define what your events mean.
It seems like your probabilities $P_i$ are not the probabilities of event $i$ occurring in the sample space including all events. In a two event system, rather than $P_1=P(E_1)$, I think $P_1$ ought to be defined as the probability of event 1 happening in a world where all the other events can't happen. I.e.,
$$ P_1 = (E_1|\neg E_2), \text{ and similarly } P_2 = (E_2|\neg E_1) $$
What you want is $P(E_1 \cup E_2) = P(E_1) + P(E_2)$ because $E_1$ and $E_2$ are disjoint. By the theorem of total probability you can write $$P(E_1) = P(E_1|E_2)P(E_2) + P(E_1|\neg E_2)P(\neg E_2) = 0 + P_1(1-P(E_2)) = P_1(1-P(E_2))$$ And similarly $$P(E_2) = P_2(1-P(E_1))$$
Plugging these into the formula for one event or the other happening you get
$$\begin{align}
P(E_1 \cup E_2)
&= P(E_1) + P(E_2)\\
&= P_1(1-P(E_2)) + P_2(1-P(E_1))
\end{align}$$
Unfortunately you can't solve this for $P(E_1) + P(E_2)$ and the situation doesn't get any better as you add more events, although it does generalize easily. Nevertheless, your first step ought to be to define the problem correctly. For example you had
$P(E_2) = (1-P_1)*P_2$
whereas I think the following is correct based on my reasoning above
$P(E_2) = (1-P(E_1))*P_2$
and it's only by being very clear in the problem definition (which sadly means notation) that you can avoid subtle problems.
I've left some general comments from my original answer below in case they help.
events are not independent, specifically if one occurs all the following ones cannot occur
For the second event to happen event 1 must not have happened. So the probability of the second event is:
$P(E_2) = (1-P_1)*P_2$
This is not correct. $P(A \text{ and } B)=P(A)P(B)$ if and only $A,B$ are independent.
Question 1 ... the combined probability of exactly one event occurring will be simply the sum of the probabilities of all the events?
yes, exactly right, for any disjoint events $P(A\text{ or }B) = P(A) + P(B)$, as in Nicholas R. Peterson's comment.
(Note the general rule regardless of whether $A$ and $B$ are disjoint is $P(A\text{ or }B) = P(A) + P(B) - P(A\text{ and }B)$ which is easiest to see on a Venn diagram.)
Best Answer
The probability to have $k$ successes in $n$ independent trials when each success occurs with probability $1/2$ is given by $$ {1\over 2^n}\pmatrix{n \cr k} $$ where $\pmatrix{n \cr k} = {n! \over k!(n-k)!}$ is the binomial coefficient that counts the number of subsets with k elements in a set with n elements. So you want to compute $${1\over 2^{10}}\left(\pmatrix{10 \cr 7}+\pmatrix{10 \cr 8}+\pmatrix{10 \cr 9}+\pmatrix{10 \cr 10}\right) $$ My computer tells me that this is equal to $11/64 \simeq {\bf 0.17}$.
This can be also computed by hand by noting that $\pmatrix{n \cr k} = \pmatrix{n \cr n-k}$. This gives $$ \pmatrix{10 \cr 7} = \pmatrix{10 \cr 3} = {10\times 9 \times 8 \over 3\times 2 \times 1}, \quad \pmatrix{10 \cr 8} = \pmatrix{10 \cr 2} = {10\times 9 \over 2 \times 1}, \quad \pmatrix{10 \cr 9} = \pmatrix{10 \cr 1} = {10\over 1}, \quad \pmatrix{10 \cr 10} = 1. $$ The binomial coefficients can also be computed by hand using Pascal's triangle.