A collection of $n$ coins is flipped. The outcomes are independent, and the $i$-th coin comes up heads with probability $\alpha_i, i=1, \dots, n$. Suppose that for some value of $j, 1 \leq j \leq n, \alpha_j=\frac{1}{2}$. Find the probability that the total number of heads to appear on the $n$ coins is an even number.
Now, I don't understand the importance of the statement "Suppose that for some value of $j, 1 \leq j \leq n, \alpha_j=\frac{1}{2}$"
That one coin guarantees that the probability is exactly $1/2$.
If the sum from the other coins is odd and that one turns up "heads", then the total is even; otherwise it's odd.
$$ \begin{align} & {}\quad\Pr(\text{even}) = \mathbb E(\Pr(\text{even} \mid \text{other coins' outcomes})) \\[8pt] & = \sum_{\text{other coins' outcomes}} \Pr(\text{even}\mid\text{particular other coins' outcome})\cdot\Pr(\text{particular other}\ldots) \\[8pt] & = \sum_\text{outcomes} \frac12\cdot \Pr(\cdots) \\[8pt] & = \frac12,\text{ since the sum of those probabilities is }1. \end{align} $$
Slightly later edit: Just to be concrete: Suppose $\Pr(\text{1st coin H}) = 9/10$ and $\Pr(\text{2nd coin H})=1/2$. Then $$ \begin{align} \Pr(\text{even}) & = \Pr(\text{1st H})\cdot\Pr(\text{even}\mid\text{1st H}) + \Pr(\text{1st T})\cdot\Pr(\text{even}\mid\text{1st T}) \\[8pt] & = \frac9{10}\cdot\frac12 + \frac1{10}\cdot\frac12 \quad = \frac12\left(\frac9{10}+\frac1{10}\right) \quad = \frac12\cdot1. \end{align} $$
