Writing WFFs with exactly two two-place connectives from truth tables

[E92M3]

I was given a truth table and I must write a wff with exactly two two-place connective. I am new to logic and don't know where to start. I need to find wff a), b), c) and d).

C B A ... wff a)
T T T ... T
T T F ... F
T F T ... T
T F F ... F
F T T ... T
F T F ... T
F F T ... T
F F F ... T

A B C ...wff b)
T T T ... F
T T F ... T
T F T ... F
T F F ... F
F T T ... F
F T F ... T
F F T ... F
F F F ... T

I have no clue on these first two. I tried many but they all seemed not to work.

A B ... wff c)
T T ... F
T F ... T
F T ... T
F F ... T

I think I can do this one:
this is equivalent to ~(A&B) but I need to use two two place connectives so I wrote ~(A&(B&B)) is this correct? Also the question didn't specify whether I can use "~". can I use it anyway?

A ... wff d)
T ... T
F ... T

This one I can also manage but I'm not sure what is the right answer. This is equivalen to (Av~A) which I can write as ((A&A)v~A). However, I can also write ((A&A)->A) or ((A&A)<->A) aswell. I think there's a few more which is right?

 

[magwas]

The first step in the algebraic way is to write the minterm for the variables for each row where you see T as the outcome you leave all variables where the income is T as is, and negate those where the variable is F

for example
ABC
TFT = T
becomes
$$A\overline{B}C$$

For a graphical solution you can draw the Karnaugh table of the function, and try to cover the T's with some (possibly overlapping) rectangular areas. You come up with the solution in a similar way, just now you can drop inputs where they are both T and F in the rectangle.

I would say that the question did not say anything about one-place connectives, so using "~" should be okay.
You are right that there can be more solutions to such a problem.
I see no errors in your solutions above.