The first picture shows $5$ squares, the $4$ smaller ones, along with the $2\times 2$ square. Counting $1\times 1$, $2\times 2$, and $3\times 3$ squares, the second picture shows $14$ squares.
With this reasoning, a $4\times 4$ grid would equal $30$.
In general, an $n\times n$ grid would equal $1^2+2^2+\ldots+n^2=\frac{n(n+1)(2n+1)}{6}$.
These are the square pyramidal numbers, because they are the number of cubes needed to build a pyramid with a square base $n$ levels high.
$5$ clues is the minimum to force a unique solution on a $5 \times 5$ board. You showed that there is indeed a way to force a unique solution with $5$ clues, and below is a proof that you cannot force a unique solution with less than $5$ clues.
If you have $4$ clues, and two of the clues are in the same row (or column), then there will be two rows (or two columns) that have no clues, and those rows (columns) can be swapped for any solution to obtain another solution. So, if you could do it with only $4$ clues, you definitely want all clues to be in different rows and columns.
Likewise, if you have two clues with the same number (or color), then there will be two numbers (or colors) that are not used any of the clues, and hence you can swap those numbers (colors) for any solution to obtain another solution. So, if you could do it with only $4$ clues, you definitely want all clues to be of all different numbers and all different colors.
Without loss of generalization, we can therefore say that the clues are $1A$, $B2$, $C3$, and $D4$, and also without loss of generalization we can assume the clues are placed as follows (remember that with the clues in all different rows and columns, we can keep swapping any two rows and columns to end up in this configuration):
\begin{array}{|c|c|c|c|c|}
\hline
A1&.&.&.&.\\
\hline
.&B2&.&.&.\\
\hline
.&.&C3&.&.\\
\hline
.&.&.&D4&.\\
\hline
.&.&.&.&.\\
\hline
\end{array}
So, if we can find wo solutions to this puzzle, then we know that $4$ clues can never force a unique solution. And indeed there are two solutions:
\begin{array}{|c|c|c|c|c|}
\hline
A1&E3&D2&B5&C4\\
\hline
E4&B2&A5&C1&D3\\
\hline
D5&A4&C3&E2&B1\\
\hline
B3&C5&E1&D4&A2\\
\hline
C2&D1&B4&A3&E5\\
\hline
\end{array}
\begin{array}{|c|c|c|c|c|}
\hline
A1&E4&D5&B3&C2\\
\hline
E3&B2&A4&C5&D1\\
\hline
D2&A5&C3&E1&B4\\
\hline
B5&C1&E2&D4&A3\\
\hline
C4&D3&B1&A2&E5\\
\hline
\end{array}
Note that the second solution is the first solution mirrored along the diagonal with the clues (which itself can be achieved through a single rotation together with a vertical or horizontal mirroring, i.e. swapping of rows or columns). Indeed, I didn't have to provide two solutions at all to make the point that the $4$ clues as indicated cannot force a unique solution, since given that all the clues are on the diagonal, then if it has any solution at all, then it has a mirror solution as well.
This last observation partially answers your third question as well: the argument I gave above clearly generalizes to show that every $n \times n$ puzzle of this kind will require at least $n$ clues to force a unique solution: with $n-1$ clues, they have to be, without loss of generalization, all along the diagonal, and hence if there is any solution at all, there will always be another.
OK, but can you always force a unique solution with exactly $n$ clues? That is still an open question ... we know it works for $n=5$, but frankly I doubt you can do it for $n>5$.
As far as your second question goes, I found four valid boards that cannot be obtained from one another through swapping colors, numbers, row, columns, or doing any rotation or mirroring:
\begin{array}{|c|c|c|c|c|}
\hline
A1&B3&C4&D5&E2\\
\hline
D4&A2&B5&E3&C1\\
\hline
E5&D1&A3&C2&B4\\
\hline
B2&C5&E1&A4&D3\\
\hline
C3&E4&D2&B1&A5\\
\hline
\end{array}
\begin{array}{|c|c|c|c|c|}
\hline
A1&B3&C4&D5&E2\\
\hline
D3&A2&E5&C1&B4\\
\hline
E4&C5&A3&B2&D1\\
\hline
B5&E1&D2&A4&C3\\
\hline
C2&D4&B1&E3&A5\\
\hline
\end{array}
\begin{array}{|c|c|c|c|c|}
\hline
A1&B3&C2&D5&E4\\
\hline
C4&A2&E5&B1&D3\\
\hline
B5&D4&A3&E2&C1\\
\hline
E3&C5&D1&A4&B2\\
\hline
D2&E1&B4&C3&A5\\
\hline
\end{array}
\begin{array}{|c|c|c|c|c|}
\hline
A1&B3&C2&D5&E4\\
\hline
C5&A2&D4&E3&B1\\
\hline
B4&E5&A3&C1&D2\\
\hline
E2&D1&B5&A4&C3\\
\hline
D3&C4&E1&B2&A5\\
\hline
\end{array}
I am pretty sure that all other valid boards can be transformed into one of these $4$ through swapping colors, numbers, row, columns, or doing any rotation or mirroring. For example, the two earlier boards can be seen to be of the third type by putting all the $A$'s along he diagonal in order, followed by a diagonal mirroring. So, I am pretty sure the answer to your second question is $4$.
Best Answer
The left column's entries should be interpreted with padded zeroes on the right, so that e.g. the last summand is 100,000,000.