Probability that no consecutive heads or tails occur in a sequence in which exactly 10 tails occurred and the last outcome is a tail

I am solving a problem given in Hugh Gordon's "Discrete Probability" book (Section 3.1, Problem 10)

Problem

A coin is tossed repeatedly until tails has occurred ten times.

a. What is the probability that, at that point, heads has not occurred twice in a row?

b. What is the probability that, at that point, tails has not occurred twice in a row?

My attempt of a solution and reasoning

Note:the answer at the back of a book is: a. 0.056314; and b. 0.001953

a. We know that each sequence of exactly 10 tails should end with a tail and that any sequence with exactly 10 tails and 11 or more heads necessarily has at least one occurrence of the HH. Therefore our task devolves to finding all the sequences with up to 10 tails, with the restriction that they must be put in between the 10 heads (but not at the end!). Such are the sequences of (10 heads, 0 tails), (10 heads, 1 tails), ..., (10 heads, 10 tails). Each one of those sequences can be constructed in $(^{10}_n)$ ways and has a probability of $(\frac{1}{2})^{10+n}$.
Hence, the total probability is just an addition of all the cases, i.e., $p=\sum_{i=0}^{10}(^{10}_i)\cdot (\frac{1}{2})^{10+i}\approx .0563135147095$, which agrees with the answer provided by the author.

b. Here we are dealing with almost the opposite situation as compared to (a) since there is an infinite number of sequences that have exactly ten tails but unlimited amount of heads interspersed between them (again, no head can occur after the last tail). These sequences are (10 tails, 10 heads), (10 tails, 11 heads), (10 tails, 12 heads),...,(10 tails, n heads),..., and each of those can be generated by selecting one new spot for every additional "H" among the 10 spots "in between", i.e. $(^{10}_1)$ , while the probability of occurrence for each of those sequences is $(\frac{1}{2})^{20+n}$. Hence, $p=(^{10}_1)\cdot \sum_{i=0}^{\infty}(\frac{1}{2})^{20+i}$, which does not correspond to the answer provided by the author. What am I missing?

a. The probability of seeing $HH$ before $T$ is $0.25$, and the probability of not seeing $HH$ before $T$ is $0.75$. We are looking for the probability of not seeing $HH$ before $T$ for each of $10$ tails. $$P=0.75^{10}\approx0.0563135147$$

b. The probability of seeing $H$ before $T$ is $0.5$, and the probability of not seeing $H$ before $T$ is $0.5$. We are looking for the probability of seeing $H$ before $T$ for each of $9$ tails. (We do not need to see $H$ before the first $T$). $$P=0.5^{9}=0.001953125$$