Alternately, you could fully build the tree. What is the probability that you picked a fair coin? What is the probability that it shows heads five times in a row? The unfair coin? Five times in a row?
And don't forget conditional probability.
It sounds like you already have the intuition since you understand that the answer is obtained by dividing the number of outcomes with exactly 3 heads by the total number of outcomes. From here it's a matter of understanding how to calculate these two things.
The total number of outcomes is simply $2^6 = 64$ since we're tossing a coin 6 times and each toss has only two possible outcomes.
The number of outcomes with exactly 3 heads is given by ${6 \choose 3}$ because we essentially want to know how many different ways we can take exactly 3 things from a total of 6 things. The value of this is 20.
So the answer is $20/64 = 5/16$.
The error you made is thinking that "number of outcomes with exactly 3 heads" is equal to "half of the total number of outcomes of 6 tosses." If this were the case then logically, "exactly 3 tails" must also be exactly half of the total outcomes. This means that "exactly 3 heads or exactly 3 tails" must describe all possible outcomes (because each scenario joined by the "or" would have probability $1/2$) but this is clearly not the case since we can have, e.g., 1 heads and 5 tails, etc.
To put it another way... You said any sequence is equally likely. That is correct. But sequences containing exactly 3 heads do not make up half of the total number of sequences. Therefore it does not follow that the probability is $1/2$. The easiest way to see this clearly is to list every possible outcome. But for 64 outcomes it can be tedious, so let's do it with a simpler and similar problem.
Say we want to know the probability of getting exactly 2 heads if we flip a coin 4 times. Unless I'm misunderstanding your misunderstanding, your earlier thinking would lead you to believe the answer is $1/2$. But if we list all possible outcomes:
HHHH
HHHT
HHTH
HHTT
HTHH
HTHT
HTTH
HTTT
THHH
THHT
THTH
THTT
TTHH
TTHT
TTTH
TTTT
We see that only 6 of them have exactly 2 heads. $6/16 = 3/8$. And if we do this problem the way I answered the original one, then the total number of outcomes is $2^4 = 16$ and the total number of outcomes with exactly 2 heads is ${4 \choose 2} = 6$. So we again get $6/16 = 3/8$.
Best Answer
The probability that you get no tails when you flip a fair coin $10$ times is $\left(\frac12\right)^{10}$. The probability that you get at least one tail is therefore $1-\left(\frac12\right)^{10}$. The probability that each of the $1000$ coins comes up tails at least once is then $$\left(1-\left(\frac12\right)^{10}\right)^{1000}\approx0.37642\;.$$ We want the probability of the complementary event, which is therefore about $0.62357$.
As a quick crude estimate note that $$\left(1-\left(\frac12\right)^{10}\right)^{1000}\approx\left(1-\frac1{1000}\right)^{1000}\approx\frac1e\approx0.37\;.$$