In general, if $A$ and $B$ are events, then the conditional probability of $A$, given that $B$ has happened, (in symbols, $\Pr(A|B)$), is defined by the formula
$$\Pr(A|B)=\frac{\Pr(A\cap B)}{\Pr(B)}.$$
In our case, we have $B$ is the event "at least $1$ head." Note that $\Pr(B)=\frac{7}{8}$.
To find the probability of exactly $1$ tail, given that there is at least $1$ head, let $A$ be the event "exactly $1$ tail." Then $A\cap B$ is also the event "exactly $1$ tail," and, as you calculated, has probability $\frac{3}{8}$. It follows that the conditional probability of exactly $1$ tail, given there is at least one, is, by the displayed formula above, equal to
$$\frac{\frac{3}{8}}{\frac{7}{8}},$$
which simplifies to $\frac{3}{7}$.
Or we can think of the events HHH, HHT, HTH, HTT, THH, THT, and TTH as equally likely, given there is at least one head. So each has (conditional) probability $\frac{1}{7}$. Now for the conditional probabilities we can count. I do not advocate this approach, since the defining formula is far more versatile.
In the case of your random variable $T$, it turns out to have values $0$, $1$, or $2$, with probabilities $\frac{1}{7}$, $\frac{3}{7}$, and $\frac{3}{7}$ respectively. Note that $T$ is an ordinary random variable, but not the same random variable as the (unconditional) number of heads, since we are disregarding any experiment in which we got $0$ heads. In particular, the probabilities add to $1$. (The probabilities you computed did not.)
Now we calculate the cumulative distribution function $F_T(t)$, which, as you know, is $\Pr(T\le t)$.
If $t\lt 0$, then $F_T(t)=0$.
If $0\le t\lt 1$, then $F_T(t)=\frac{1}{7}$.
If $1\le t\lt 2$, then $F_T(t)=\frac{4}{7}$.
Finally, if $t\ge 2$, then $F_T(t)=1$.
Let's do some substitutions first do make this look a little nicer: If we let $k=x-r, n=r, m=X-n$ and $q=1-p$, the identity can be written as
$$\sum_{k=0}^m \binom{n+k-1}{k}q^k=\sum_{k=0}^m \binom{m+n}{m-k}q^{m-k}(1-q)^{k}$$
and changing $k \to m-k$ on the RHS, we obtain the equivalent
$$\sum_{k=0}^m \binom{n+k-1}{k}q^k=\sum_{k=0}^m \binom{m+n}{k}q^k(1-q)^{m-k}$$
Now, both sides are polynomials in $q$ of degree $m$, so it suffices to compare the coefficients.
The coefficient of $q^s$ of the LHS is just $\binom{n+s-1}{s}$. On the RHS, for specific $k$, the coefficient of $q^s$ in the summand is just
$$(-1)^{s-k}\binom{m+n}{k}\binom{m-k}{m-s}$$
So we are left to prove the identity
$$\sum_{k=0}^s (-1)^{s-k}\binom{m+n}{k} \binom{m-k}{m-s}=\binom{n+s-1}{s}$$
Noting that $(-1)^s\binom{n+s-1}{s}=\binom{-n}{s}$, we see that this identity is a special case of the general identity
$$\sum_{k=0}^{\infty} (-1)^k \binom{N}{k} \binom{a-k}{b}=\binom{a-n}{b-n}$$
where $N=m+n, a=m, b=m-s$. The latter identity can be found e.g. here.
Best Answer
The statistical term for what you are looking for is the so called quantile function. For a random variable with distribution function $F$ it is defined as follows:
$$Q(p)\,=\,\inf\left\{ x\in \mathbb{R} : p \le F(x) \right\} $$
For continuous, strictly increasing distribution functions, this is equivalent to:
$$Q(p) = F^{-1}(p)$$
The modification with $\inf\{\cdot\}$ is required for cases as the one you describe, where you have a non-bijective distribution functions. This is the case for discrete distribution functions such as the binomial distribution.
Notice that you wrote that you need to find $x$ such that:
$$\alpha = \sum_{k\ =\ 0}^x\binom{n}{k}\ p^k\;\left(1-p\right)^{\ n-k}$$
In fact such $x$ does not necessarily exist. Instead you are looking for the smallest $x$ such that:
$$\alpha \leq \sum_{k\ =\ 0}^x\binom{n}{k}\ p^k\;\left(1-p\right)^{\ n-k}=:B(n,p,x)$$
A very easy (naive) way to find this value on your own (for discrete distributions as the binomial) is to calculate the above right hand side for increasing value of $x$ until you find the first $x$ satisfying your condition:
With your values for $\alpha, n ,p$ we get:
$$ \begin{aligned} B(100,0.3, 0) &\approx 3.2\cdot10^{-16} \\ B(100,0.3, 1) &\approx 1.4\cdot10^{-14} \\ &\vdots \\ B(100,0.3, 30) &\approx 0.549 \\ B(100,0.3, 31) &\approx 0.633 \\ B(100,0.3, 32) &\approx 0.711 \\ B(100,0.3, 33) &\approx 0.779 \\ \end{aligned} $$
As you can see, the first $x$ for which $B(100,0.3,x)$ exceeds $\alpha=0.7$ is $x=32$ which is also the output of the Excel formula.