These can be solved at the same time, the closed roll is just a special case of the open ended.
Let
$$d = \text{Number of sides on each die}$$
$$n = \text{Number of dice rolled}$$
$$s = \text{Score required for success}$$
$$m = \text{number of successes required}$$
Also, let
$$s = n^k+r$$
where
$$k\in 0,1,2,\dots$$
$$r\in 1,2,\dots,d$$
And, let
$$\begin{align}
p_o &= \text{chance of getting open ended}\\
&= \begin{cases}
\frac{1}{d}&,\text{if roll is open}\\
0&,\text{otherwise}\\
\end{cases}
\end{align}$$
So, for 1 die, we have a geometric distribution to get $k$ open ended results followed by a single roll to get $r$ (note that this is not exactly a geometric distribution since we need $k$ failures with $q=1-p_0$ followed by 1 success with a different probability). Also we will require $x^0=1$.
$$p_1=\left(p_0\right)^{k}\left(\frac{d-r+1}{d}\right)$$
To get $m$ success from $n$ dice, we have a binomial distribution as you noticed so
$$\begin{align}
p_{m,n}&=\sum_{i=m}^n\binom{n}{i}p_1^i\left(1-p_1\right)^{n-i}\\
\end{align}$$
Some examples:
- $d=6$, $n=1$, $s=4$, $m=1$ & open ended; so $k=0$, $r=4$
$$\begin{align}
p_0=\frac{1}{6}\\
\end{align}$$
$$\begin{align}
p_1&=\left(\frac{1}{6}\right)^{0}\left(\frac{6-4+1}{6}\right)\\
&=1\left(\frac{3}{6}\right)\\
&=1\times\frac{1}{2}\\
&=\frac{1}{2}\\
\end{align}$$
$$\begin{align}
p_{1,1}&=\sum_{i=1}^1\binom{1}{i}p_1^i\left(1-p_1\right)^{1-i}\\
&=\binom{1}{1}\left(\frac{1}{2}\right)^{1}\left(1-\frac{1}{2}\right)^{1-1}\\
&=1\times\frac{1}{2}\times{1}\\
&=\frac{1}{2}\\
\end{align}$$
- $d=6$, $n=1$, $s=10$, $m=1$ & open ended; so $k=1$, $r=4$
$$\begin{align}
p_0=\frac{1}{6}\\
\end{align}$$
$$\begin{align}
p_1&=\left(\frac{1}{6}\right)^{1}\left(\frac{6-4+1}{6}\right)\\
&=\frac{1}{6}\times\frac{3}{6}\\
&=\frac{1}{12}\\
\end{align}$$
$$\begin{align}
p_{1,1}&=\sum_{i=1}^1\binom{1}{i}p_1^i\left(1-p_1\right)^{1-i}\\
&=\binom{1}{1}\left(\frac{1}{12}\right)^{1}\left(1-\frac{1}{12}\right)^{1-1}\\
&=1\times\frac{1}{12}\times{1}\\
&=\frac{1}{12}\\
\end{align}$$
- $d=6$, $n=3$, $s=4$, $m=2$ & closed; so $k=0$, $r=4$
$$\begin{align}
p_0=0\\
\end{align}$$
$$\begin{align}
p_1&=\left(0\right)^{0}\left(\frac{6-4+1}{6}\right)\\
&=1\left(\frac{3}{6}\right)\\
&=\frac{1}{2}\\
\end{align}$$
$$\begin{align}
p_{3,2}&=\sum_{i=2}^3\binom{1}{i}p_1^i\left(1-p_1\right)^{1-i}\\
&=\binom{3}{2}\left(\frac{1}{2}\right)^{2}\left(1-\frac{1}{2}\right)^{3-2}+\binom{3}{3}\left(\frac{1}{2}\right)^{3}\left(1-\frac{1}{2}\right)^{3-3}\\
&=3\times\frac{1}{4}\times\frac{1}{2}+1\times\frac{1}{8}\times1\\
&=\frac{4}{8}\\
&=\frac{1}{2}\\
\end{align}$$
- $d=6$, $n=3$, $s=10$, $m=2$ & open ended; so $k=1$, $r=4$
$$\begin{align}
p_0=\frac{1}{6}\\
\end{align}$$
$$\begin{align}
p_1&=\left(\frac{1}{6}\right)^{1}\left(\frac{6-4+1}{6}\right)\\
&=\left(\frac{1}{6}\right)\left(\frac{3}{6}\right)\\
&=\frac{1}{12}\\
\end{align}$$
$$\begin{align}
p_{3,2}&=\sum_{i=2}^3\binom{1}{i}p_1^i\left(1-p_1\right)^{1-i}\\
&=\binom{3}{2}\left(\frac{1}{12}\right)^{2}\left(1-\frac{1}{12}\right)^{3-2}+\binom{3}{3}\left(\frac{1}{12}\right)^{3}\left(1-\frac{1}{12}\right)^{3-3}\\
&=3\times\frac{1}{144}\times\frac{11}{12}+1\times\frac{1}{1,728}\times1\\
&=\frac{34}{1,728}\\
&=\frac{17}{864}\\
&\approx0.02\\
\end{align}$$
For roll $i$, let $X_i=1$ if roll $i$ of the die is 1 and 0 otherwise. Hence,
$$R_1 = \sum\limits_{i=1}^R X_i$$
The dice cannot roll 6 in the first $R$ rolls, so $X_i=1$ with probability 1/5 and 0 with probability 4/5. Hence, $$E[R_1|R] = \sum\limits_{i=1}^R E[X_i|R] = \frac{1}{5}R$$
$$Var[R_1|R] = \sum\limits_{i=1}^R Var[X_i|R] = \frac{1}{5}\frac{4}{5}R$$
By the law of total variance,
\begin{align*}
Var[R_1] &= Var[E[R_1|R]] + E[Var[R_1|R]]\\
&= Var[\frac{R}{5}] + E[\frac{4}{25}R]\\
&= \frac{1}{25}Var[R] + \frac{4}{25}E[R]\\
&= \frac{1}{25}\frac{1-1/6}{1/36} + \frac{4}{25}5\\
&= 2
\end{align*}
Best Answer
Indeed $T$ is the number of success among $n$ trials where the probability of success is $1/6$, so it should follow the binomial distribution $\mathcal B(n,1/6)$.
More formally, let $X_i$ be the side of the $i^{th}$ roll. Then $T=\sum_{i=1}^n1_{\{X_i=X_{n+i}\}}$. The random variables $(X_i,X_{n+i})_{1\le i\le n}$ are independent, and for all $i\in\{1,\cdots,n\}$, $1_{\{X_i=X_{n+i}\}}$ equals $1$ with probability $1/6$ and $0$ with probability $5/6$ hence it follows the Bernoulli distribution $\mathcal B(1/6)$. Then $T$ is the sum of $n$ independent Bernoulli variables with parameter $1/6$, so $T$ is a Binomial variable with parameter $(n,1/6)$.
For the second question, you have $X=\sum_{i=1}^nX_i1_{\{X_i\in\{2,4,6\}\}}$. Let $Y_i=X_i1_{\{X_i\in\{2,4,6\}\}}$. Clearly the distribution of $Y_i$ is given by $$ \mathbb P(Y_i=2)=\mathbb P(Y_i=4)=\mathbb P(Y_i=6)=\frac16\quad\textrm{and}\quad\mathbb P(Y_i=0)=\frac12, $$ hence $$ \mathbb E[Y_i]=2\quad\textrm{and}\quad\operatorname{var}(Y_i)=\frac{16}{3}\cdot $$ Since the variance of the sum of independent random variables is equal to the sum of the variances, we have $$ \operatorname{var}(X)=\sum_{i=1}^n\operatorname{var}(Y_i)=\mathbb E[X^2]-\mathbb E[X]^2, $$ hence $$ \mathbb E[X^2]=\sum_{i=1}^n\operatorname{var}(Y_i)+\left(\sum_{i=1}^n\mathbb E[Y_i]\right)^2=\frac{16}{3}n+4n^2. $$