Your answers to part (a) and (b) are correct.
Your answer to (c) is not. You could get (5,5), (5,6), (6,5) or (6,6). There are two ways of scoring 11.
$$P(S \ge 10) = \left(\frac{5}{21}\times \frac{5}{21}\right)+\left(\frac{5}{21}\times \frac{6}{21}\right)+\left(\frac{6}{21}\times \frac{5}{21}\right)+\left(\frac{6}{21}\times \frac{6}{21}\right) = \frac{121}{441}$$
Part (d): I thought an ace was a playing card!
Part (e): The even numbers are 2, 4, and 6. The numbers greater than four are 5 and 6, so
$$\left( \frac{2}{21}+\frac{4}{21}+\frac{6}{21} \right) \times \left(\frac{5}{21}+\frac{6}{21}\right) = \frac{44}{147}$$
EDIT
I think that by an ace you mean at least one number 1.
The way to fail is to get all not-ones. The probability of not getting a one is 20/21. The probability of getting $n$ not-ones is $(20/21)^n$. If the probability of failure is $(20/21)^n$ the the probability of success is $1-(20/21)^n$. We need to solve $1-(20/21)^n > 1/2$ for $n$. Well:
\begin{array}{ccccccc}
1-\left(\frac{20}{21}\right)^{\!n} &>& \frac{1}{2} &\iff& \frac{1}{2} & > & \left(\frac{20}{21}\right)^{\!n} \\ \\ \\\
&&&\iff& \log\left(\frac{1}{2}\right) &>& n\log\left(\frac{20}{21}\right) \\ \\ \\
&&&\iff& \frac{\log(1/2)}{\log(20/21)} &<& n
\end{array}
Hence $n > 14.2$, meaning that $n \ge 15$. We need to roll the dice at least 15 times to have more than a 50-50 chance of getting at least a single ace.
Best Answer
The die is loaded, so the probability of a single die rolling a six is $\frac {36}{91}$.
[Because $1/91$ is the constant of proportionality determined by the sum of weights (i.e. $\sum\limits_{j=1}^{6} j^{2} = 1^2{+}2^2{+}3^2{+}4^2{+}5^2{+}6^2 = 91$) and $36$ is the weight for result of a single die to be
6.]Further, the probability of obtaining the third six on the seventh roll is the probability six occurs exactly twice somewhere within the first six rolls, and once on the seventh: $$\dbinom{6}{2}\dfrac{(36)^3(55)^4}{91^7}\approx0.1239{\small 3}$$