Are Cantor type numbers algebraic? This question may be naive. Take an infinite set of distinct algebraic numbers (hence countable). List them out in a table (randomly) by picking a choice of ordering and change the diagonal numbers. 
1) Is it possible to decide if this Cantor type diagonal number is algebraic or transcendental?
2) For some choice of ordering the table, can one prove there exists a Cantor type algebraic or transcendental number?
3) Can we have infinitely many Cantor type transcendental numbers from an infinite subset of choice of orderings?
 A: If one uses binary representation (so that changing the digit means swapping 0 and 1), then every real arises as a diagonal real. If you want the diagonal to be $d$, start with any list of algebraic numbers, and first change the diagonal digits on the list to be the dual to $d$. This is a finite change to each of the reals on the list, so they remain algebraic, but now diagonalizing the new list will produce $d$. 
(One should say a bit more about how the diagonal procedure handles the integer part, the sign and the binary point, but there are reasonable procedures that allow this argument to go through.)
Meanwhile, of course, one of the main points of Cantor's construction was that if you put all algebraic reals on the list, then the diagonal is definitely transcendental. 
Finally, you use the word "decide" in the question, which suggests that you might be proposing it as a problem of computational decidability. That is, the question would be: given a program that enumerates a list of algebraic reals, by giving more and more of their digits, can one compute yes-or-no whether the diagonal real for this list is algebraic? The answer to this is no, this is not computable. The reason is that we can design devious enumeration programs. Namely, suppose that $f$ was a computable function that decided the nature of the diagonal of any given c.e. enumeration of algebraic numbers. By the recursion theorem (which enables this kind of circular-seeming argument), we may design a program $p$ such that at first, the program $p$ appears to be enumerating a list with all $1$s on the diagonal, so that the diagonal number would be $0$. It does this until $f(p)$ gives its output, which must say that the number is algebraic (since otherwise we would refute it by continuing with our pattern). But after we see this output given for $f(p)$, then program $p$ deviously shifts to a mode where it forces the diagonal real to be transcendental, thereby refuting $f$. So there can be no such computable procedure $f$ to determine whether the diagonal is algebraic.
The question of whether a given c.e. listing of algebraic reals has an algebraic diagonal seems to have complexity $\Pi^0_2$ in the index of the list, and I believe it is complete at this level of complexity. 
A: To answer the third question in the affirmative, consider a list containing all algebraic numbers enumerated according to Dedekind for instance. Then the diagonal is transcendental. (Therefore we can use binaries without excluding certain periodic representations.) Each of the permutations of the list will also result in a transcendental diagonal. However, these diagonals need not all be different. But in order to obtain infinitely many transcendentals, we can bolster the list by inserting some algebraic numbers repeatedly. This will result in infinitely many transcendental diagonals.
