Interesting question: From a deck of 52 cards pick 26 at

random is better

Random is not better, both give equal win rates according to simulation.

No difference, you can think of the first 26 cards in the shuffled deck as the randomly selected 26 cards, and then you pick the first two. So the winning probability will be exactly the same.

First option is slightly better.

One way to argue if the 26 random cards are even, then its the same as the 2nd situation, but if its uneven (12R 14B), then probability is 12/26 * 11/25 + 14/26 * 13/25 > the probability of the 2nd option. And the probability just gets better as the draw is more skewed.

Another way to argue is the 1st option is picking from a deck of 52 even cards, and 2nd option is picking from 26. First option probability is 25/51, second option is 12/25. As u choose from more and more cards, the probability increases and tends towards 1/2.

Intuitive Solution.
26 is an arbitrary selection. For a two card case:
Case 2: deck of two cards. 1 black and 1 white card gives P = 0.
Case 1: pick two cards from 52. P > 0.

Random is better. Here is the solution:

For latter game(fixed 26 cards with equal red and black), the probability to win is:
p2=1-13*13/C(2,26)=0.48. (1-probability of picking two different color cards)

For the random game, although the expectations of number of red card and black card are equal, but they may not be the exactly same. Assume R is the number of red cards, and B is the number of black cards. with constraints: R+B=26
then the probability to win for this game becomes:
p1=1-R*B/C(2,26)

With the constraints R+B=26, then R*B = p2 always.

fix is better, since in random case more colors are mixed in. The prob of hitting same color pair got lowered. Keep in mind a deck of cards is composed of 4 colors and each of 13 cards. So that 12+14 is not happening.