a. Peter can increase the number of marbles by $20$. The first three times he should exchange $20$ blue marbles for $28$ red ones. He will then have $111 - 60 = 51$ blue marbles and $111 + 3 \cdot 28 = 111 + 84 = 195$ red ones. Then, he should exchange $11$ red marbles for $7$ blue ones. He will end up with $184$ red and $58$ blue marbles. He will have $242$ marbles altogether, which is $20$ more than the initial $222$.

b. Exchanging $11$ red marbles for $7$ blue marbles decreases the total amount of marbles by $4$. Exchanging $20$ blue marbles for $28$ red marbles increases the total amount of marbles by $8$. The initial amount of marbles is even, so the total will remain even after any number of exchanges. Hence, the total number of marbles cannot be increased by $33$ since this number is odd.

c. Assume that this can be done. Let $x$ denote the number of red marbles. Then there are $3x$ blue marbles and the total number of marbles is $4x$. After each exchange the number of marbles increases or decreases by a multiple of $4$. Since there were $222$ marbles at the beginning, only the numbers of the form $222 + 4k$ can be reached. The equation $222 + 4k = 4x$ has no integer solutions since $222$ is not divisible by $4$. Hence, Peter can never have three times as many blue marbles as red.