Let $P_n$ be the probability that after $n$ trials the number of successes is even.
Let $p$ be the probability of success on any one trial. In our problem, $p=2/3$, but we might as well generalize a bit.
The number of successes after $n+1$ trials can be even in two ways: (a) After $n$ trials we had an odd number of successes and we got a success on the $(n+1)$-th trial; or (b) After $n$ trials we had an even number of successes, and we had a failure on the $(n+1)$-th trial. The probability of (a) is $p(1-P_n)$ and the probability of (b) is $(1-p)P_n$. We therefore have the recurrence
$$P_{n+1}=p(1-P_n)+(1-p)P_n=p+(1-2p)P_n.\qquad (\ast)$$
The recurrence $(\ast)$ is linear, and there are general tools for solving such recurrences. But the recurrence is particularly simple, as is the physical situation, so we will use a trick.
It is intuitively clear that if $p(1-p)\ne 0$ and $n$ is large, then $P_n$ should be close to $1/2$. Let $P_n=1/2+y_n$, and substitute in $(\ast)$. There is a lot of cancellation, and we obtain
$$y_{n+1}=y_n(1-2p). \qquad (\ast\ast)$$
Note that $y_0=1/2$, since if there are $0$ trials, for sure there are $0$ successes. Each time we increment $n$ by $1$, $y_n$ gets multiplied by $1-2p$. So the sequence $(y_n)$ is the geometric sequence
$y_n=\frac{1}{2}(1-2p)^n$,
and therefore
$$P_n=\frac{1}{2}(1+(1-2p)^n).$$
If $p=0$ or $p=1$, $P_n$ is completely determined by the parity of $n$. Suppose now that $p\ne 0$ and $p\ne 1$. Then $|1-2p|<1$, so
$(1-2p)^n$ approaches $0$ as $n \to\infty$. Thus $P_n$ indeed has limit $1/2$.
Comments: $1$. The recurrence approach can be used with more complicated problems, such as determining the probability that the number of successes after $n$ trials is a multiple of $3$.
$2$. One can also use an algebraic approach which is basically a rewording of the solution by Didier Piau. Let $F_n(t)=(tp +(1-p))^n$. Expand $F_n(t)$ using the Binomial Theorem. Evaluate $F_n(t)$ at $t=1$ and at $t=-1$, add up. Suppose that $k$ is odd. Then the terms $\binom{n}{k}p^k(1-p)^{n-k}$ and $\binom{n}{k}(-p)^k(1-p)^{n-k}$ cancel. Suppose that $k$ is even. Then the terms $\binom{n}{k}p^k(1-p)^{n-k}$ and $\binom{n}{k}(-p)^k(1-p)^{n-k}$ are equal. It follows that
$$P_n=\frac{1}{2}(F_n(1)+F_n(-1))=\frac{1}{2}(1^n+(1-2p)^n).$$
$3$. When we let $y_n=1/2+F_n$, the recurrence simplified. Consider the general recurrence $P_{n+1}=a+bP_n$, where $b \ne 1$. Make the substitution $F_n=w+y_n$. We get $y_{n+1}=by_n +a+ bw-w$. By setting $w=a/(1-b)$, we get the recurrence $y_{n+1}=by_n$, which is simple to solve.
Let $N=100$. The number of steps since the last success occurred performs a Markov chain on $\{1,2,\ldots,N\}$ where each transition $k\to1$ has probability $k/N$ and each transition $k\to k+1$ has probability $1-k/N$. Successes are visits to the state $1$.
By the classical one-step analysis, the stationary measure $(\pi_k)$ of this Markov chain solves the system
$$\pi_{k+1}=(1-k/N)\pi_k\ (1\leqslant k\lt N),\qquad\pi_1=\sum\limits_{k=1}^N(k/N)\pi_k.$$ Thus,
$$
\pi_1=\frac{N^N}{N!}\,\left(\sum_{k=0}^{N-1}\frac{N^k}{k!}\right)^{-1}.
$$
Using Stirling's formula to estimate the prefactor and the central limit theorem to estimate the parenthesis, one sees that, when $N\to\infty$,
$$
\pi_1\sim\sqrt{\frac2{\pi N}},
$$
hence, for $N$ fixed, when $T\to\infty$, the number of visits to the state $1$ up to time $T$ is approximately
$$
\sqrt{\frac2{\pi N}}\,T.
$$
Assuming that $N$ is large enough to be considered as a large number of steps for the Markov chain on $\{1,2,\ldots,N\}$ (a rather dubious hypothesis), the mean number of visits of state $1$ (aka, the mean number of successes) would scale as
$$
\sqrt{\frac{2N}\pi},
$$
which, for $N=100$, is about $7.98$.
Best Answer
Your second question is much easier. Number the balls $1$ to $100$, so balls $1$ to $10$ are red. If you never replace any balls, then your probability space consists of all sequences of $n$ numbers, each between $1$ and $100$, with no repeats. The number of such sequences is $100\cdot 99\cdots (100-n+1)=\frac{100!}{(100-n)!}$. A successful sequences consists of $x$ red balls and $n-x$ other balls in some order. The number of successful sequences is $\binom{10}x\cdot \binom{90}{n-x}\cdot n!$ (choose $x$ red balls, choose $n-x$ non-red balls, then order them). Therefore, the probability of success is $$ \frac{\binom{10}x\cdot \binom{90}{n-x}\cdot n!}{\frac{100!}{(100-n)!}}=\frac{\binom{10}x\cdot \binom{90}{n-x}}{\binom{100}{10}} $$ This is the hypergeometric distribution.
When you have "partial replacement," so red balls are kept and non-reds are returned, then there is no simple formula. Imagine that instead of stopping after $n$ draws, you continue until all red balls are drawn. Let $T_1$ be number of draws to get your first red ball, let $T_2$ be the number of draws it takes to get your second, and so on up to $T_{10}$. Then $T_k$ is a geometric random variable for each $k$, with probability of success $(10-(k-1))/(100-(k-1))$. That is, $$ P(T_k=m) = (1-p_k)^{m-1}p_k,\qquad \text{where }p_k=\frac{11-k}{101-k} $$ You want to find the probability that after $n$ draws, you have exactly $x$ red balls. In order for this to occur, you need to have drawn your $x^{th}$ red ball before drawn number $n$, which means that $T_1+\dots+T_x\le n$. However, you also need to not have drawn any more red balls before draw $n$, which is equivalent to saying $T_1+\dots +T_x+T_{x+1}> n$. In order words, we want to compute $$ P(T_1+\dots+T_x\le n)-P(T_1+\dots+T_x+T_{x+1}\le n) $$ A good tool for computing independent sums of discrete random variables is probability generating functions. The probability generating function for a geometric distribution $Z$ with probability of success $p$ is $$ G_{Z}(s):=\sum_{i\ge 0}P(Z=i)s^i=\frac{sp}{1-(1-p)s} $$ Furthermore, the p.g.f. for the sum of random variables is the product of their p.g.f's. Finally, we can recover the cumulative density function from a random variable $Z$ by extracting the coefficient of $x^i$ in $\frac{G_Z(s)}{1-s}$. That is, $$ P(Z\le i)=\text{coefficient of $s^i$ in } \frac{G_Z(s)}{1-s} $$ Putting this altogether, we get
\begin{align} P(\text{$x$ red balls in $n$ draws}) = \text{coefficient of $s^n$ in }\frac1{1-s}\left(\prod_{k=1}^x\frac{p_ks}{1-(1-p_k)s}\right)\left(1-\frac{p_{x+1}s}{1-(1-p_{x+1})s}\right) =\text{coefficient of $s^n$ in }\frac1{1-(1-p_{x+1})s}\left(\prod_{k=1}^{x}\frac{p_ks}{1-(1-p_k)s}\right) \end{align} This is difficult to evaluate by hand, but can be done easily with a computer if $x$ and $n$ are small enough. The following Mathematica code does this: