Probability of obtaining a heads on the coin before a 1 or 2 on the die?

A simple way to find the probability is to condition on the result of the first round. It is clear there is some probability $p$ of obtaining a head before (though not necessarily immediately before) a $1$ or $2$. Call that the probability of winning.

We win if (i) we get a head on the first round or (ii) we get a tail, don't roll a $1$ or $2$, but ultimately win.

The probability of (i) is $\frac{1}{2}$.

For (ii), note that the probability of tail and then something other than $1$ or $2$ is $\frac{1}{2}\cdot \frac{4}{6}$. Given this has happened, the probability of ultimately winning is $p$. Thus $$p=\frac{1}{2}+p\cdot \frac{1}{2}\cdot \frac{4}{6}.$$ Solve this linear equation for $p$. We get $p=\frac{3}{4}$.

This is the probability of no-heads and probability of no-1,2 before than a head appear in $n$ play, where a play is tossing in order first a coin and after a dice.

So a play before the last play (when a head happens) is no-head AND no-1,2. Because the two events are independent one of each other (coin and dice) then we have that the probability for some $n$ that a head happen before a 1 or 2 in the dice is

$$\left(\frac12\cdot\frac46\right)^{n-1}\cdot\frac12$$

because the probability that the dice show something different than one or two is $\frac46$, and the probability than the coin show tail or a head is $\frac12$. Then we have $n-1$ plays where we cant have a head or a 1 or 2, and in the last play we can have in the coin a head (in the dice doesnt matter what we get after we toss the coin).

Then the probability that this happen in any $n$ number of plays is the probability that this happen in one play OR two plays OR three plays OR..., i.e.

$$\sum_{n\ge 1}\left(\frac13\right)^{n-1}\cdot\frac12=\frac12\sum_{n\ge 0}\left(\frac13\right)^n=\frac12\cdot\frac1{1-\frac13}=\frac34$$

Use recursion; Let $p$ be probability of your event. Then, we have $$p= \frac{1}{2} + \frac{1}{2}\times\frac{2}{3}\times p,$$ where first term is probability of having head in first toss and second term results from tail in coin toss and 3-6 in first roll and having head before 1-2 in the next tries.

Thus, $p=0.75$.