a) and b): We count the rows of the chessboard from the top. The $n \times n$ chessboard has $\frac{n+1}{2}$ rows with odd numbers, and in each of these rows, there are $\frac{n+1}{2}$ black squares. Each of these $\left(\frac{n+1}{2}\right)^{2}$ black squares must be covered by a different triomino, so at least $\left(\frac{n+1}{2}\right)^{2}$ triominoes are required. Since each consists of 3 squares, due to the no-overlap rule, it must hold that: $3\left(\frac{n+1}{2}\right)^{2} \leq n^{2}$. For 1, 3, and 5, this gives $3 \leq 1$, $12 \leq 9$, $27 \leq 25$. Therefore, for $n<7$, the required covering is not possible.
The figure shows, as an anchor, a covering for $n=7$ with $\left(\frac{n+1}{2}\right)^{2}=16$ triominoes and the step $n \rightarrow n+2$, which proves the existence of a covering for every $n \geq 7$. In this step, an "L" shape, 5 fields wide and high, is added to the right bottom corner of the $n \times n$ chessboard, whose covering with $\left(\frac{n+1}{2}\right)^{2}$ stones is assumed, and then completed with $2 \times 2$ fields, each with a matching triomino. The "L" contains 5 triominoes, and the two $2 \times (n-5)$ rectangles together contain $n-3$ triominoes. Thus, there are $\left(\frac{n+1}{2}\right)^{2} + n - 3 + 5 = \frac{n^{2} + 6n + 9}{4} = \left(\frac{n+3}{2}\right)^{2}$ triominoes, the minimum number for the $(n+2) \times (n+2)$ chessboard. By complete induction, the claim follows.
Note: Only all black squares should be covered; white squares can remain uncovered. Of course, the triominoes do not extend beyond the chessboard. A complete solution also includes proving that the added triominoes in the step $n \rightarrow n+2$ are compatible with the minimum number, as well as mentioning $n=1$.
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