Let X count the winnings:
X = 0 if none of the coins land heads up: P(X = 0) = 1/16.
X = 1 if only the penny lands heads up: P(X = 1) = 1/16.
X = 5 if only the nickel lands heads up: P(X = 5) = 1/16.
X = 6 if only the nickel and the penny land heads up: P(X = 6) = 1/16.
X = 10 if only one of the dimes land heads up--there are two ways for this to happen, so P(X = 10) = 1/8.
X = 11 if one dime and the penny both land heads up--there are two ways for this to happen, so P(X = 11) = 1/8.
X = 15 if one dime and the nickel both land heads up--there are two ways for this to happen, so P(X = 15) = 1/8.
X = 16 if one dime, the nickel and the penny all land heads up--there are two ways for this to happen, so P(X = 16) = 1/8.
X = 20 if both dimes land heads up: P(X = 20) = 1/16.
X = 21 if both dimes and the penny land heads up: P(X = 21) = 1/16.
X = 25 if both dimes and the nickel land heads up: P(X = 25) = 1/16.
X = 26 if all the coins land heads up: P(X = 26) = 1/16.
B.) I assumed that each of the coins was fair and that the result of the tossing of one coin is independent of the results of the tossing of any of the other coins.
C.) We take the probabilities of each of the values of X times those values of X and then find the sum of those products:
1/16 * 0 + 1/16 * 1 + 1/16 * 5 + 1/16 * 6 + 1/8 * 10 + 1/8 * 11 + 1/8 * 15 + 1/8 * 16 + 1/16 * 20 + 1/16 * 21 + 1/16 * 25 + 1/16 * 26 = 208/16 = 13, so a contestant can expect to win 13 cents on average if the game is played a large number of times.