Solution:

a.
It is not possible. Indeed, let $C_{1}, \ldots, C_{m}$ be the colored squares, let $a_{i}$ be the number of colored squares adjacent to $C_{i}$ for $i=1, \ldots, m$, and let $l$ be the number of sides shared by two adjacent colored squares. We have $a_{1}+\cdots+a_{m}=2l$, since every side is shared by two squares and is therefore counted twice; in particular, $a_{1}+\cdots+a_{m}$ is an even number. If $m$ were odd and $a_{i}$ were odd for every $i$, then the sum $a_{1}+\cdots+a_{m}$ would be odd, and thus we would have a contradiction.

b.
It is possible. Consider, for example, the four squares located in the first two rows and the first two columns of the chessboard, and the $2 \times 2$ square formed by the squares located in the second and third rows and the second and third columns of the chessboard. The square in the second row and second column is common to the two squares and is the only one that has four adjacent colored squares. All the other colored squares have 2 adjacent colored squares.

c.
It is not possible. Indeed, suppose we color some squares of the chessboard in such a way that each of them has 2 or 4 adjacent colored squares. Let us distinguish the squares of the chessboard into two types, according to what their normal black-and-white coloring would be. To avoid confusion, let us suppose that the coloring required by the problem is done in green. Let us call $b$ the number of green squares of the first type and $n$ the number of green squares of the second type. Moreover, let us write $b=b_{2}+b_{4}$, where $b_{2}$ is the number of squares of the first type with two adjacent green squares and $b_{4}$ is the number of squares of the first type with 4 adjacent green squares; similarly, let us write $n=n_{2}+n_{4}$. Since the squares adjacent to those of the first type are necessarily of the second type, and vice versa, by counting the number $l$ of sides shared by two green squares as in part (a), we obtain $l=2b_{2}+4b_{4}=2n_{2}+4n_{4}$, from which $b_{2}-n_{2}=2\left(n_{4}-b_{4}\right)$ is necessarily an even number. But then $b_{2}+n_{2}=\left(b_{2}-n_{2}\right)+2n_{2}$ is also necessarily even, and therefore a coloring with the properties required in (c) is impossible.

Alternative:

c.
It is not possible. To prove this, let us count in two different ways the number $l$ of sides shared by two colored squares.

(i) Denoting by $c_{2}$ the number of colored squares with 2 adjacent colored squares, and by $c_{4}$ the number of colored squares with 4 adjacent colored squares, we will have that
$$l=\frac{2c_{2}+4c_{4}}{2}=c_{2}+2c_{4},$$
where the division by 2 accounts for the fact that in the numerator each side has been counted twice. If $c_{2}$ were odd, it would follow from this formula that $l$ would also have to be odd.

(ii) Let us now distinguish the squares of the chessboard into two categories, according to what their standard black-and-white coloring would be. To avoid confusion, let us suppose that the coloring required by the problem is done in green. Since a pair of adjacent green squares always consists of a former-white square and a former-black square, it follows that in order to compute $l$ it suffices to sum all the adjacencies starting from former-white green squares, without dividing by 2. But since from every green square there start either 2 or 4 adjacencies, this sum is necessarily even, in contrast with the previous point.