Your approach using generating functions is correct. I will work the whole solution so others still trying to understand generating functions can follow.
First, we know the probability will be the following occurrences,
( 8 tails and 17 heads with runs < 6)/( 8 tails and 17 heads)
To calculate the function occurrences without runs visualize it as a boxes and balls problem. Separate the boxes with the 8 Tails
_T_T_T_T_T_T_T_T_
This leads to 8 boxes and 17 balls to place. (25 - 8 = 17)
Since our boxes can contain 0-5 balls the function takes the form for a single box
$x^0+x^1+x^2+x^3+x^4+x^5=\mathbf{1+x^1+x^2+x^3+x^4+x^5}$
Since we have 9 boxes, we take this result to the 9th power
$\mathbf{(1+x^1+x^2+x^3+x^4+x^5)^9}$
using $\frac{1-x^{m+1}}{1-x} = 1+x+x^2+...+x^m $
$$\mathbf{1+x^1+x^2+x^3+x^4+x^5 =\frac{(1-x^6)^9}{(1-x)^9}}$$
Break this into two functions f(x) and g(x)
$$f(x)= (1-x^6)^9$$
$$f(x)= \frac{1}{(1-x)^9}$$
Their corresponding expansions are,
$(1-x^m)^n=1-\binom{n}{1}x^{1*m}+\binom{n}{2}x^{2*m}....+(-1)^n\binom{n}{n}x^{nm}$
$\frac{1}{(1-x)^n}=1+(\frac{1+n-1}{1})x+(\frac{2+n-1}{2})x^2+......+(\frac{r+n-1}{r})x^r$
We need to find the coefficient for $x^{17}$ which is equal to
$$\Sigma f(x)g(x)$$
Where the resulting power of x is 17
The only values of $\mathbf{f(x)}$ that can meet this constraint are $\mathbf{x^0,x^6,x^{12}}$
Lets represent the power of x in f(x) with $a^k$ and g(x) with $b^k$
This leads to the only valid combinations being
$$a^0b^{17}+a^6b^{11}+a^{12}b^5$$
Therefore, the number of occurrences of 8 tails and 17 heads with runs < 6 is
$$1*\binom{17+9-1}{17}-\binom{9}{1}\binom{11+9-1}{11}+\binom{9}{2}\binom{5+9-1}{5}$$
This evaluates to 447669 occurrences.
Next we find all possible occurrences with no constraints.
Using the formula for placement of n identical objects in k boxes $\binom{n+k-1}{n}$
Which gives us
$$\binom{17+9-1}{17}=1081575$$
$$100*\frac{447669}{1081575}\approx41.39\%$$
