In the second case, your pair can be any one of $(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)$ and you'd satisfy "obtaining a pair".
That gives you six possible pairs, each of which one has probability of occurring $\dfrac 1{36}$ gives us $$6\times \frac 1{36} = \frac 16$$
Now, if you want to know what the probability of rolling two dice simultaneously and obtaining two sixes (one prespecified pair of the six possible pairs), that would be $\dfrac 1{6\cdot 6} = \dfrac 1{36}$.
With this distinction made, yes, the probability of obtaining two sixes when rolling one die twice, and the probability of rolling two sixes simultaneously are equal.
You can use generating functions here.
Suppose we represent a single die with the function $x + x^2+x^3+x^4+x^5+x^6$
The different exponents here represent the different possible outcomes of rolling the die.
Now, if you want the different possible outcomes of rolling two dice, just multiply this function by itself:
$$(x + x^2+x^3+x^4+x^5+x^6)\cdot(x + x^2+x^3+x^4+x^5+x^6)$$
$$= x^2+2x^3+3x^4+4x^5+5x^6+6x^7+5x^8+4x^9+3x^{10}+2x^{11}+x^{12}$$
Again, the different exponents here represent the possible outcoems, so they range from $2$ to $12$
But what's really neat, is that the coefficients represent the number of ways to get that outcome.
For example, the term $5x^6$ indicates that there are $5$ ways to get an outcome of $6$
Why does this work? Well, we get one term $x^6$ by multiplying $x$ from the first term and $x^5$ from the second (i.e. by throwing a $1$ with the first die, and a $5$ with the second), but we also got a term $x^6$ by multiplying $x^2$ from the first term and $x^4$ from the second (i.e. by throwing a $2$ with the first die, and a $4$ with the second), etc.
OK, so for $6$ dice, we just need to find the different coefficients and exponents for the following function:
$$(x + x^2+x^3+x^4+x^5+x^6)^6$$
Here is where a tool like WolframAlpha comes in:
Scroll down to the expanded form, and you'll find:
$$x^{36} + 6 x^{3}5 + 21 x^{34} + 56 x^{33} + 126 x^{32} + 252 x^{31} + 456 x^{30} + 756 x^{29} + 1161 x^{28} + 1666 x^{27}$$
$$ + 2247 x^{26} + 2856 x^{25} + 3431 x^{24} + 3906 x^{23} + 4221 x^{22} + 4332 x^{21} + 4221 x^{20} + 3906 x^{19}$$
$$ + 3431 x^{18} + 2856 x^{17} + 2247 x^{16} + 1666 x^{15} + 1161 x^{14} + 756 x^{13} + 456 x^{12} + 252 x^{11} + 126 x^{10}$$
$$ + 56 x^9 + 21 x^8 + 6 x^7 + x^6$$
So, for example, given the term $2856x^{17}$ we know that there are $2856$ ways for the $6$ dice to add up to $17$, meaning that the probability of getting a sum of $17$ is $\frac{2856}{46656}$
Best Answer
Presumably, the question is asking about a 99% chance of rolling a 6 at some point in the $n$ rolls.
André's comment is the most straightforward approach. The complement of the event "rolling a 6 [at some point]" is "never rolling a 6," and it is much easier to compute this quantity.
You can complete your first approach to recover the same answer.
Let $p=1/6$ be the probability of rolling a $6$. With your first method, the geometric series can be rewritten as $$p(1+(1-p) + (1-p)^2 + \cdots + (1-p)^{n-1}) = p \frac{1-(1-p)^n}{1-(1-p)} = 1-(1-p)^n.$$
For this quantity to be $ \ge 0.99$, you need $(1-p)^n < 0.01$.
Your second approach is flawed: the probability you have computed is the probability that you roll a 6 exactly once and on a particular roll (e.g. the last roll).