Regarding your first question, note that a subset has an equivalent representation as a binary sequence. For example, for $n = 4$, the subset $\{1,3\}$ can be identified with $(1,0,1,0)$. Now, picking a subset uniformly at random, is like picking a sequence uniformly at random out of the $2^n$ possibilities. You should be able to argue the rest.
Regarding your 2nd question, let $X_i$ be the indicator that the $i$ is present in the random subset (i.e., $i$-th position in the binary representation above has a $1$) Then, you want
$$
E\Big[ \sum_{i=1}^n i X_i \Big| \sum_{i=1}^n X_i = k\Big] = \sum_{i=1}^n i\, E\Big[X_i \Big| \sum_j X_j = k\Big]
$$
Now, you can use symmetry. Let $a_i :=E[X_i \Big| \sum_j X_j = k]$ (which is the conditional probability of choosing the $i$). Then all $a_i$ should be equal and $\sum_i a_i = k$ (why?), from which it follows that $a_i = k/n$. This give you the answer that you have, and also shows that the conditional probability of choosing the $i$ is $k/n$ for every $i$.
A "different" why to see that: we can call the chosen of some number of the set as some random variable $X_i$, and we can define the random variable of the sum as
$$X=\sum_{i=1}^{k}X_i$$
Then we have that
$$\Bbb E[X]=\Bbb E\left[\sum_{i=1}^{k}X_i\right]=\sum_{i=1}^{k}\Bbb E[X_i]$$
But we have that the $X_i$ are not independent but anyway they expected value is the same because
$$\Bbb E[X_i]=\sum_{x_1,x_2,...,x_i}x_i\Pr[X_i=x_i|X_1=x_1,X_2=x_2,...,X_{i-1}=x_{i-1}]\cdot\Pr[X_1=x_1,X_2=x_2,...,X_{i-1}=x_{i-1}]$$
where
$$\Pr[X_i=x_i|X_1=x_1,X_2=x_2,...,X_{i-1}=x_{i-1}]=\begin{cases}\frac1{n-i+1}\quad&\text{if }x_i\notin\{x_1,...,x_{i-1}\}\\0\quad&\text{if }x_i\in\{x_1,...,x_{i-1}\}\end{cases}$$
and
$$\Pr[X_1=x_1,X_2=x_2,...,X_{i-1}=x_{i-1}]=\frac1{(n)_{i-1}}$$
where $(a)_b$ is a falling factorial. So for any $x_i$ value we will have a limited number of (i-1)-tuplas $(x_1,x_2,...,x_{i-1})$ where probability is not zero. The number of these (i-1)-tuplas is $(n-1)_{i-1}$.
Then we have
$$\Bbb E[X_i]=\sum_{k=1}^{n}\frac{(n-1)_{i-1}}{(n)_{i-1}}k\frac{1}{n-i+1}=\frac1n\sum_{k=1}^{n}k=\frac{n+1}{2}$$
Then finally
$$\Bbb E[X]=\sum_{j=1}^{k}\Bbb E[X_j]=k\frac{n+1}{2}$$
Best Answer
There are no subsets of $10$ elements with $1,2,3...,9$ as the maximum element. The number of subsets with $m(\ge10)$ as the maximum element is given by $\binom{m-1}9$, since you select $9$ other numbers from the $m-1$ numbers smaller than $m$. There are $\binom{20}{10}$ subsets of size $10$ in totality. The required expectation is$$\frac1{\binom{20}{10}}\sum_{m=10}^{20}m\binom{m-1}9=\frac1{9!\cdot\binom{20}{10}}\sum_{m=10}^{20}\frac{m!}{(m-10)!}$$Now use the fact
$\frac{m!}{(m-10)!}=m(m-1)...(m-9)\\=\frac{(m+1)-(m-10)}{11}[m(m-1)...(m-9)]\\=\frac1{11}[(m+1)m...(m-9)-m(m-1)...(m-10)]$
and so$$\sum_{m=10}^{20}\frac{m!}{(m-10)!}=\frac{21\times20\times19\times...\times11}{11}=21\times20\times19\times...\times12$$giving the required expectation as $210/11$.
In general, when subsets of size $q$ of the set $\{1,2,...,n\}~(q\le n)$ are taken and their maximum noted, the expected maximum is$$\frac{(n+1)q}{q+1}$$