Two-space notchers not fully connected
So long as a is not occupied by White, Black can live (provided Black plays first). The vital point is at b, defending the cutting point inside.
Black can live
Black has to be careful in responding to the hane at . Only playing at
will live. Playing at a is wrong.
Playing at is wrong. White makes a placement tesuji at
, causing an exchange at
, followed by connecting at
. Black faces a shortage of liberties - being unable to play at a - and dies because the eye on top cannot be formed.
is a very careless move.
is a tesuji, followed by the atari at
which destroys an eye in sente, and then
destroys another eye. Now Black is dead.
is also wrong. White plays a placement at
, followed by cutting at
in sente (aiming to capture three black stones at a), before making the final killing move at
.
The descent at is incorrect as well. White can kill the group with a hane at
(aiming to capture two black stones at a), followed by a placement at
.
attempts to divide the space to make two eyes, but the connection at
leaves Black with a shortage of liberties, and Black cannot play at b to defend the second eye.
White can kill
If White starts at b, Black's 222WN notcher dies (2 and 3 are miai). Either hane (5 or the other one) kills as well (White gets the second hane or nakade). White can also kill from outside by filling (222WW).
Starting with 2 would let Black live (with 1). Therefore fails if
is one to the right (222NN), but hane still kills. Actually hane kills any 222xy.
If the lower leg is longer (223WN), White can not kill (hane is met with 1, and 1 with 2). Neither if the upper leg is longer (223NW, similar).
Two wedges kill
The two stones make it impossible for the black group to live, even though it has extra liberties outside. Try figuring it out yourself, using the last three diagrams in the "Black can live" section as reference.
The solution can be found at flying saucer shape (222WW).