Your approach b) is wrong: both the single step updating, in which all data are used together to update the prior and arrive at the posterior, and the Bayesian sequential (also called recursive) updating, in which data are used one at a time to obtain a posterior which becomes the prior of the successive iteration, must give exactly the same result. This is one of the pillars of Bayesian statistics: consistency.
Your error is simple: once you updated the prior with the first sample (the first "Head"), you only have one remaining sample to include in your likelihood in order to update the new prior. In formulas:
$$P(F|HH) =\frac{P(H|H,F)P(F|H)}{P(H|H)} $$
This formula is just Bayes' theorem, applied after the first event "Head" has already happened: since conditional probabilities are probabilities themselves, Bayes' theorem is valid also for probabilities conditioned to the event "Head", and there's nothing more to prove really . However, I found that some times people don't find this result self-evident, thus I give a slightly long-winded proof.
$$P(F|HH) =\frac{P(HH|F)P(F)}{P(HH)}= \frac{P(H|H,F)P(H|F)P(F)}{P(HH)}$$
by the chain rule of conditional probabilities. Then, multiplying numerator and denominator by $P(H)$, you get
$$\frac{P(H|H,F)P(H|F)P(F)}{P(HH)}=\frac{P(H|H,F)P(H|F)P(F)P(H)}{P(HH)P(H)}=\frac{P(H|H,F)P(H)}{P(HH)}\frac{P(H|F)P(F)}{P(H)}=\frac{P(H|H,F)}{P(H|H)}\frac{P(H|F)P(F)}{P(H)}=\frac{P(H|H,F)P(F|H)}{P(H|H)}$$
where in the last step I just applied Bayes' theorem. Now:
$$P(H|H,F)= P(H|F)=0.5$$
This is obvious: conditionally on the coin being fair (or biased), we are modelling the coin tosses as i.i.d.. Applying this same idea to the denominator, we get:
$$P(H|H)= P(H|F,H)P(F|H)+P(H|B,H)P(B|H)=P(H|F)P(F|H)+P(H|B)P(B|H)=0.5\cdot0.\bar{3}+1\cdot0.\bar{6}$$
Finally:
$$P(F|HH) =\frac{P(H|H,F)P(F|H)}{P(H|H)}=\frac{0.5\cdot0.\bar{3}}{0.5\cdot0.\bar{3}+1\cdot0.\bar{6}}=0.2$$
QED
That's it: have fun using Bayesian sequential updating, it's very useful in a lot of situations! If you want to know more, there are many resources on the Internet: this is quite good.
1&2) You are working on a prediction problem. You need to solve $$\pi(k\ge450|4\text{ success of 10 tries})=1516927277253024\sum_{k=450}^{1000}\int_0^1\binom{1000}{k}p^k(1-p)^{1000-k}(1-p)^{25}p^{23}\mathrm{d}p$$
That is your posterior probability that the next 1000 tosses will result in 450 or more successes. You should gamble no more than the prize times the probability.
3) No, choosing the prior that you believe to be closest to the truth is the most scientific solution. You should use any real information that you have. You can test the sensitivity of your result to your prior, but since you are gambling what you need to do is work out the knowledge that you really have.
If you believe it is a nearly fair coin then $B(20,20)$ as a prior is quite reasonable, though $B(2,2)$ is as well.
EDIT
You are solving $$\pi'(k=K|X)=\int_0^1f(k=K|p)\pi(p|X)\mathrm{d}p$$ for each value of $k\ge{450}$, where $f$ is the likelihood function, $\pi$ is the posterior and $\pi'$ is the prediction. You are removing the uncertainty regarding $p$ by marginalizing it out. You have to sum all the cases in your hypotheses space which is $k\ge{450}$.
You are looking at the likelihood of every outcome weighted by the posterior probability over the entire set of all possible parameter values.
Best Answer
Yes, it does converge to the "true distribution" (suitably defined)
First of all, it is worth noting that it is a little strange to refer to the "true distribution" of the parameter as something aside from the prior and posterior. If you proceed under the operational Bayesian approach then the parameter has an operational definition as a function of the sequence of observable values, and so it is legitimate to refer to a "true value" of the parameter (see esp. Bernardo and Smith 2000). The standard convergence theorems in Bayesian statistics show that the posterior converges weakly to the true parameter, defined operationally through the law-of-large numbers. It is less common to refer to a "true distribution" of the parameter, as something apart from the prior or posterior. If I understand your intention correctly, that would essentially just be a point-mass distribution on the true value. If that is what you mean, then yes, the posterior will converge (weakly) to this. That is really just another restatement of the standard convergence theorems in Bayesian statistics.
Taking $\mathbf{X} = (X_1, X_2, X_3, ...)$ to be the sequence of observable coin-toss outcomes, we define $p_H$ operationally as a function of $\mathbf{X}$. In Bayesian analysis with IID data the most operational definition of the parameters is an index to the limiting empirical distribution of the observable sequence (see O'Neill 2009 for discussion and details). Now, the beta posterior you are referring to arises in the IID model:
$$X_1,X_2,X_3,... | p_H \sim \text{IID Bern}(p_H).$$
The parameter $p_H$ can be given an operational definition as the limit of the sample mean of the observable coin-tosses (see O'Neill 2009 again). To facilitate our analysis, we will use the notation $\hat{p}_H \equiv \lim N_H/N$ to denote this limit, so the "true value" of the parameter $p_H$ is the point $\hat{p}_H$. (In other words $p_H$ is $\hat{p}_H$, but we will use two separate referents to elucidate the convergence.)
Your posterior distribution does indeed converge to the "true distribution" of $p_H$, which is a point-mass distribution on $\hat{p}_H$. To see this, we first derive the asymptotic mean and variance$^\dagger$ of the posterior:
$$\begin{equation} \begin{aligned} \lim_{N \rightarrow \infty} \mathbb{E}(p_H| \mathbf{X}_N) &= \lim_{N \rightarrow \infty} \frac{a_0+N_H}{a_0 + b_0 + N} \\[6pt] &= \lim_{N \rightarrow \infty} \frac{N_H}{N} \cdot \frac{a_0+N_H}{N_H} \Bigg/ \frac{a_0+b_0+N}{N} \\[6pt] &\overset{a.s}{=} \lim_{N \rightarrow \infty} \frac{N_H}{N} = \hat{p}_H. \\[6pt] \lim_{N \rightarrow \infty} \mathbb{V}(p_H| \mathbf{X}_N) &= \lim_{N \rightarrow \infty} \frac{(a_0+N_H)(b_0+N_T)}{(a_0 + b_0 + N)^2(a_0 + b_0 + N+1)} \\[6pt] &= \lim_{N \rightarrow \infty} \frac{a_0+N_H}{a_0 + b_0 + N} \cdot \frac{b_0+N_T}{a_0 + b_0 + N} \cdot \frac{1}{a_0 + b_0 + N + 1} \\[6pt] &\leqslant \lim_{N \rightarrow \infty} \frac{1}{a_0 + b_0 + N + 1} \\[6pt] &= 0. \\[6pt] \end{aligned} \end{equation}$$
So, we have $\mathbb{E}(p_H| \mathbf{X}_N) \overset{a.s}{\rightarrow} \hat{p}_H$ and $\mathbb{V}(p_H| \mathbf{X}_N) \rightarrow 0$, which gives convergence in mean-square to the true parameter value $\hat{p}_H$. Using Markov's inequality this implies convergence in probability to $\hat{p}_H$, which further implies convergence in probability of the posterior to the point-mass distribution on $\hat{p}_H$. This means that we have weak convergence to the "true distribution" of the parameter.
$^\dagger$ Your stated posterior variance is incorrect - I have used the correct posterior variance in my working.