Not sure what the real question is, but the water level will get lower when the anchor is thrown overboard (h’ < h).
When the anchor is in the boat the water replacement depends on its mass and when the anchor is in the water it depends on it’s volume.

You need to cut 3 links.
After these cuts you have 3 pieces of length 2, 4, 8 and a remaining piece of length 4
And also 3 broken links (2+4+8+4+3×1 = 21)

With one broken link and the pieces of length 2, 4 and 8 we can count to 15 in binary.
For numbers above 15 we can use the remaining 6 links as an offset.

There are 3 Dutch teachers but Peter isn’t one of them.
So Karl, Simon and Steven teach Dutch.
Mr. Dutch doesn’t teach Dutch because Karl does.
So Peter is Mr. Dutch.
Mr. English teaches history and his name isn’t Simon.
Steven teaches Dutch and chemistry so he isn’t Mr. English.
The only one left is Karl, so Karl is Mr. English and he teaches Dutch and history.
Which means that Peter Dutch teaches maths and chemistry.
Mr. Painter doesn’t teach maths or chemistry (Mr. Dutch) so he can’t be Steven and must be Simon.
Which leaves Steven to be Mr. Writer.

Simon Painter teaches Dutch and history
Peter Dutch teaches maths and chemistry
Steven Writer teaches Dutch and chemistry
Karl English teaches Dutch and history

The dragon brings water from the 7^th well.
The knight expects this so before the dual he drinks from one of the other wells and thus poisons himself, but by drinking from the dragons glass he will be cured.
The knight doesn’t bring any water from the wells but some pure not poisoned water.
The dragon thinks it is poisoned so he drinks from well 7 trying to cure himself but in fact he poisons himself without possibility to be cured.

The ratio of boys to girls will stay the same

When the chance of having a girl is G, the chance of having a boy is (1-G).

The chance of having only a boy is (1-G)
The chance of having a girl and a boy is(1-G)G
The chance of having 2 girls and a boy is (1-G)(G^2)
The chance of having N girls and a boy is (1-G)(G^N)

Now we have to multiply the probability with the number of boys/girls and add them together the get the average number of boys/girls.

For boys we get the following sum: (1-G) + (1-G)G + (1-G)(G^2) + …. + (1-G)(G^N) + …
This is a sum in the form of A + AB + AB^2 +  … + AB^N where B < 1.
When N goes to infinity the sum goes to A/(1 – B). Filling in A=(1-G) and B=G results in (1-G)/(1-G) = 1.

For girls we get (1-G)G + 2(1-G)(G^2) + … + N(1-G)(G^N) + … = (1-G) x (G + 2G^2 + 3G^3 + … + NG^N + …).
Here we have the sum of n*X^n which results in x/(x-1)^2.
The average number of girls is (1-G) (G / (G -1)^2) = G(1-G)/(G-1)(G-1) = -G/(G-1) = G/(1-G)

The average number of kids is 1 + G/(1-G) = 1/(1-G)

The percentage of girls is (G/(1-G)) / (1/(1-G)) = G(1-G)/(1-G) = G.

So the ratio of boys to girls will stay the same.

$18.56

3 x $18.56          = $55.68
$56.18 – $0.50 = $55.68

When the 5 cards are picked at random, this is only possible when the “neat” pile isn’t that neat.

The suit is easy. With five cards there is at least one suit with 2 or more cards. Pick one of those cards and put the other one at the top. The value can be coded in binary, horizontal means 0 and vertical means 1. With 4 cards you can code up to the number 15. So the pile passed to Peter will contain cards  arranged both horizontally and vertically.

When all the cards needs to be arranged vertically (or horizontally) there is no easy solution possible (if there is any solution at all). When you use 1 card for the suit, you only have 3 cards to indicate the value. In the worst case those 3 cards could have the same value (e.g. 3 aces) and there are only 6 ways to order those 3 cards. And there are 12 possible values of the card Alex has.

FFFPAA ......

PA>

FFF..A ...PA.

<P

FFFP.A ....A.

PA>

FFF... ...PAA

<P

FFFP.. ....AA

FF>

F..P.. .FF.AA

<FA

FF.PA. ..F..A

FP>

F...A. .FFP.A

<FA

FF..AA ..FP..

FF>

....AA FFFP..

<P

...PAA FFF...

PA>

.....A FFFPA.

<P

...P.A FFF.A.

PA>

...... FFFPAA

Where F are Federation Officers, A are aliens and P is the alien pilot.

He let the clock run backwards.

Yes,
take one piece and open all 3 links (3x$1 = $3)
With those 3 open links you can connect the 4 remaining pieces together. (3x$3 = $9)
Total cost $12 ($3 + $9)