Why is not the answer to all probability questions 1/2.

To talk of probability, you need both the sample space and a probability assigned to each element of the sample space.^*

So in the case of a fair coin, the probability is given by the data that the sample space is $\{H, T\}$ and that $\Pr(H) = \frac12$ (and therefore $\Pr(T) = \frac12$). It's only because of this that you can say that the probability of heads is $\frac12$; you can't conclude this just from "either it will be heads or not".

For example, you could have a biased coin, where the sample space is still $\{H, T\}$, but $\Pr(H) = 0.9$ (and $\Pr(T) = 0.1$). In that case, you cannot use the reasoning that "either it will be heads or not" to conclude that heads and tails have equal probability.

In the case of a fair coin, the full reasoning you use is "either it will be heads, or not, and both are equally likely", where the final part (in emphasis) comes from the data given to you, namely that $\Pr(H) = \Pr(T) = \frac12$.

Even in the case of dice, you can still say "either it will be $2$ or not", but you're missing the latter part of the reasoning, that both events are equally likely. (Indeed, for fair dice, the probability of rolling $2$ is only $\frac16$, whereas the probability of rolling something other than $2$ is $\frac56$. But you could have loaded dice too.)

Often, in your textbooks, you may be given just the sample space without the probabilities, leaving implicit the fact that each individual element of the sample space has the same probability. But ideally that should be specified too.

[^*]: Note: The discussion above is of a discrete probability space. In general, we have a set ("sigma algebra") of events, and probabilities on them, satisfying certain axioms. But let's not worry about that now.

In the case of a fair coin, the two outcomes are equally probable. $P(\text{heads}) = P(\text{tails}) = 1/2$.

In the case of a (fair) dice, the two outcomes $"2"$ and $"\text{not }2"$ are not equally probable. There are five possible ways of not getting 2, all equally probable, and only one possibility to get 2. In total there are six possible outcomes. Simply put: $$\frac{5}{6} = P(\text{not }2) > P(2) = \frac{1}{6}$$.