Fastest search algorithm in a sorted list with certain error rate-limiting constraints
For simplicity of analysis I will say that 'too high' is the error condition, which is just the problem inverted, $64^{64}-n$. I also assume that requests are instant.

It takes some thinking, but waiting is never an optimal strategy. So we view the problem as just a series of guesses, done in blocks of 13 (unless we hit the 90 second timeout). Instead of timeouts our penalty will be losing guesses, and instead of minimizing time we will minimize guesses used. I write $V_2$ for the value of a guess (in bits) when we are in the state where we guessed too high twice in this block. Similarly I will write $V_1, V_0$. I write $g \in \{0\dots13\}$ for the number of guesses left in the block, including the current one.

I will evaluate a strategy where we pick three numbers $p_0$, $p_1$, $p_2$, where each $p$ encodes how skewed the binary search is for when we've guessed $0$, $1$ or $2$ wrong times in this block.

The expected value (in bits) of a guess $p \in [0, 1]$ (where $0$ is the lowest guess and $1$ is the highest) is the binary entropy function $H(p) = -p \log_2 p - (1 - p)\log_2(1-p)$.

However, for $V_2(g)$, if we guess wrong we lose the rest of this block and the next two blocks, $B$ is the value of a block in bits.

$$V_2(g) = H(p_2) - 2p_2B - p_2\sum_{k=1}^{g-1}V_2(k)$$

If we write $E_2(g)$ as the value in bits of being in state $2$ with $g$ guesses left:

$$E_2(g) = \sum_{k=1}^{g} V_2(k) = \sum_{k=1}^g \big( H(p_2) - 2p_2B - p_2E_2(k-1)\big ) $$
$$E_2(g) = g(H(p_2) - 2p_2B) - p_2 \sum_{k=1}^g E_2(k-1)$$

This recurrence can be solved:

$$E_2(g) = (H(p_2)/p_2 - 2B)(1-(1 - p_2)^g)$$

Similarly, if we guess wrong for $V_1$, we lose our advantage from state $1$:

$$V_1(g) = H(p_1) - p_1(E_1(g-1) - E_2(g-1))$$
$$E_1(g) = \sum_{k=1}^{g} V_1(k) = \sum_{k=1}^{g} \big( H(p_1) - p_1(E_1(k-1) - E_2(k-1))\big )$$

$$E_1(g) = g(H(p_1) - p_1E_2(g-1)) - p_1\sum_{k=1}^{g}E_1(k-1)$$

$$E_1(g) = (H(p_1)/p_1 - E_2(g-1))(1-(1 - p_1)^g)$$

And $E_0$ is directly analogous. Now, since $B = E_0(13)$ we have:

$$E_0(g) = (H(p_0)/p_0 - E_1(g-1))(1-(1 - p_0)^g)$$
$$E_1(g) = (H(p_1)/p_1 - E_2(g-1))(1-(1 - p_1)^g)$$
$$E_2(g) = (H(p_2)/p_2 - 2E_0(13))(1-(1 - p_2)^g)$$

Oh boy... With a CAS I found:

$$E_0(13) =\frac{1}{2P_0P_1P_2 + 1} \left(\frac{P_0H(p_0)}{p_0} - \frac{P_0P_1H(p_1)}{p1} +\frac{P_0P_1P_2H(p_2)}{p_2}\right)$$

where $P_0 = 1-(1-p_0)^{13}$, $P_1 = 1-(1-p_1)^{12}$ and $P_2 = 1-(1-p_2)^{11}$. This is the expression we want to maximize.

Numerically  maximizing this expression gives $E_0(13) \approx 6.72$ bits with:

$$p_0 \approx 0.1854365,\quad p_1 \approx 0.1422816,\quad p_2 \approx 0.0000899$$

This means that in order to get $\log_2(64^{64}) = 384$ bits we expect to need 57.1 blocks, or 1714.3 seconds. 



This isn't even optimal yet, though. $p_0, p_1, p_2$ should actually be functions of $g$ (consider that if you are on your last guess in state $0$, then you can be really greedy and not get punished). I will see if I can improve my answer.