Given, current state = q and next state = Q
a) In S-R f/f, the two inputs S and R denotes Set and Reset respectively. So, a "1" in S and "0" in R is used to set the f/f, similarly, a "0" in S and "1" in R is used to Reset the f/f(i.e. set it to 0). The input S = 1 and R = 1 is not used, since after that the f/f is in an undetermined state.
So, on input S = 1 and R = 0, we have Q = 1 (irrespective of previous state).
b) In J-K f/f, we have J and K instead of S and R. So, we have a similar truth table, except now the invalid input in SR f/f (S = 1 and R = 1), is now a valid input and is used to toggle the state of f/f.
On J = 0 and K = 0, we have Q = q (i.e. no change in f/f state). (see the table)
c) We have J = 1 and K = 1 to toggle the f/f state. (see the table)
d) In negative-edge triggered f/fs, the state of the f/f changes only on the falling edge (or negative edge) of the clock.
In the diagram above, the triangle with a bubble (small circle) in the clock input indicates that it is negative edge triggered. The absence of that bubble will indicate positive edge triggered.
e) same as b part.