There are 4 possibilities: boy-boy, boy-girl, girl-boy an girl-girl.
When they say they have a girl we are left with 3 possibilities: boy-girl, girl-boy an girl-girl.
Only one of them has 2 girls so the probability that the other child is also a girl is 1/3.

I agree with you that the person posting questions should mark answers as the correct answer.
And when there are no correct answers he should give the answer (after 2 weeks or so) because a lot of people want to know what the right answer is.
And when he doesn’t know the right answer himself he should mention that in the question.

What I am missing in the question is what bills are available in the US currency.
Of course I can look it up but it should be mentioned in the question.
Obviously the bills are not the same as for the Euro currency, because the Euro doesn’t have any 1 Euro bills only coins.

First man 1 and 2 cross the bridge, it will take them 2 minutes.
Then man 1 goes back in 1 minute.
Next man 3 and 4 cross the bridge in 10 minutes.
Man 2 returns taking 2 minutes.
And last, man 1 and 2 cross in 2 minutes.
total 2+1+10+2+2=17 minutes.

When you unwind the vine you basically get a right-angled triangle with a base of 7 * 0.9144 meter = 6.4008 meter and a hight of 6.096 meter.
The length of the vine is sqrt(6.4008^2 + 6.096^2) = 8.8388 meter (29 feet)

I doubt it is possible without cheating or be very lucky at guessing.

The friend knows 60% of the answers. For the remaining 40% of the questions he has a 50% change to guess right. So his score would be 80%.

When you guess all the answers you have a 50% change to guess it right. And that will result in a score of 50%.
With 3 cities there is a way to guarantee you have one of the three questions right, but that method also guarantees you have at least one answer wrong. For the last question you have a 50% change. And this will also result in a score of 50%.

I doubt there is a method that guarantees you have more answers right than wrong.

On then the question “Are you the jackal?” the lion and the jackal will both answer “no”.
If the parrot is in the cage behind the lion or the jackal he will also answer “no”.
If the parrot is in the cage behind the giraffe they both will answer “yes” or they both will answer “no”.
The hedgehog now knows the location of the giraffe so they didn’t all answer “no”.
If there is only one “yes” and it came from the first cage the hedgehog still doesn’t know who is in the first cage because then the first cage could contain either the parrot or the giraffe.
If there were two cages with “yes” the hedgehog would not only know the location of the giraffe but also the location of the parrot or he doesn’t know the location of the giraffe at all.
So there is only one “yes” and it is coming from the cage with the giraffe and it isn’t the first cage.

On the question “are you the giraffe?” the lion will answer “no” and the jackal will answer “yes”.
The giraffe will answer the previous question so his answer will also be “no.”
If there were two cages with “yes” the hedgehog would either know the location of the jackal and the location of the parrot or he doesn’t know the location of the jackal at all.
So there is only one cage that answers “yes” and that is coming from the cage with the jackal.

On the question “are you the parrot?” the jackal would always answer “yes” so it wouldn’t be wise to ask this to cage one if you know the jackal was in there.
We already know the first cage doesn’t contain the giraffe so only the lion and the parrot are left.
The lion would answer “no”, so he isn’t in the first cage.

The first cage contains the parrot.

Because the parrot answered “yes”, the last answer on the previous question would be “yes”.
The only animal that gave a “yes” on the previous question was the jackal.

The jackal is in the last cage

The giraffe is in the cage that answered “yes” on the first question.

The lion is in the remaining cage.

Although the hedgehog knows all the locations, we don’t.
For us there are two possibilities:

PLGJ with answers NNYN NNNY Y
PGLJ with answers NYNN NNNY Y

(The parrot answered the first question randomly with “no” and the giraffe with “yes”.)

The first riddle is simple.
All statements are mutually exclusive.
If statement 20 would be true all other statements would be false.
Which means that there can only be one true statement and 1999 false statements.
So statement 1999 “Exactly 1999 statements on this list are false.” is true.

replace “exactly” with “at least”
If statement X is true then there are at least X false statements.
Which means that there are also at least X-1 statements false and that makes statement X-1 also true.
So then X statements are true and at least X statements are false.
Because the total number of statements is 2000, X will be 1000.
The statements 1 to 1000 are true and 1001 to 2000 are false.

replace “exactly” with “at most”
Statement 2000 is true because with only 2000 statements there can be at most 2000 false statements.
If statement X and all statements after that are true then there are at most X-1 false statements which means that statement X-1 is also true.
This means that all statements before X are also true.
So all statements are true. (no false statements meets the condition of “at most X are false”.)

replace “false” with “true”
The statements are still all mutually exclusive.
So only one statement is true and that is the first statement.

replace “false” with “true” and “exactly” with “at least”
If statement X is false then all statements after X must be false because there are at most X-1 true statements.
That also means that when statement X is true all statements before statement X must be true.
And here have a problem.
If we assume that statements 2 to 2000 are all false and only look at statement 1.
If statement 1 is true than there is “at least one true statement”
But if statements 1 is false there is not “at least one true statement”
Both are solutions of the riddle.
So there are 2001 good solutions.

replace “false” with “true” and “exactly” with “at most”
If statement X is true then there are at most X true statements.
Which means that there are also at most X+1 statements true and that makes statement X+1 also true.
So there are at most X true statements but also at least (2001 – X) true statements.
The statements 1 to 1000 are false and 1001 to 2000 are true.

When a 3 digit number is added to a 4 digit number and the result is a 5 digit number then the highest digit of the 4 digit number must be 9 and the highest to of the result must be 10.
Y=9
H=1
E=0
O=4 (given)
Y+O=9+4=13. There cannot be a carry from O+U. If there was a carry A would also be 4.
A=3
Because O+U doesn’t have a carry the result is less than 10.
It also can’t be 9 because then R would be 9 or 0 and those numbers are already used.
So U must be less than 5 and the only number remaining is 2.
U=2
R can either be 6 when there is no carry from U+R or 7 when there is.
Because U=2 there will be no carry with either number.
R=6
T=U+R=8

H+A+T+E = 1+3+8+0 = 12

This is not a real riddle but just a trick of the mind where the word “of” is skipped when counting the letter “f”.