This is only an answer to your first question.
How can they replace only one entry with q and say that this is entropy of q?
In the paper $h(q)$ is not computed this way. The inequality of Lemma 4.2 is used to prove that $h(p) \le log(n)$ and
$h(p) \lt log(n)$ if $p$ is not the uniform distribution with $p_1=p_2=\ldots p_n=\frac{1}{n}$
Lemma 4.2:
$$-\sum_{i=1}^{n}p_i \log{p_i} \le -\sum_{i=1}^{n}p_i \log{q_i} \tag{1} $$
Equality holds iff $$p_i=q_i, i=1,\ldots , n \tag{2}$$
$\square$
We know that the entropy is defined by
$$h(p)=-\sum_{i=1}^{n}p_i \log{p_i} \tag{3} $$
This can be used to reformulate the inequation of the Lemma as
$$ h(p)\le -\sum_{i=1}^{n}p_i \log{q_i} \tag{4} $$
This is valid for all discrete distributions so also for the uniform distribution with
$$q_i=\frac{1}{n} ,i=1,\ldots,n \tag{4a} $$
Substituting $\frac{1}{n}$ for $q_i$ gives
$$ h(p)\le \sum_{i=1}^{n}p_i \log{n} = (\log{n}) \cdot \sum_{i=1}^{n}p_i = \log{n} \tag{5} $$
But $log{(n)}$ is also $h(q)$, if $q$ is the uniform distribution. This can checked simply by using the definition of the entropy:
$$h(q)=-\sum_{i=1}^{n}q_i \log{q_i}=-\sum_{i=1}^{n}\frac{1}{n} \log{\frac{1}{n}} = \log{n} \sum_{i=1}^{n}\frac{1}{n} = \log{n} \tag{6} $$
So it follows that for the uniform distribution $q$
$$h(p) \le \log{n} = h(q) \tag{7} $$
Because of $(6)$ and $(2)$ equality holds exactly if $p$ is the uniform distribution too.
Edit:
Theorem 5.1 states, that the continous probability density on [a,b] with $\mu = \frac{a+b}{2}$ that maximizes entropy is uniform distribution $q(x)=\frac{1}{b-a}, x \in [a,b]$. This complies with the principle of indifference for coninous variable found here.
On the whole real line there is no uniform probability density. On the whole real line there is also no continous probability density with highest entropy, because there are continous probability densities with arbitrary high entropies, e.g. the gaussian distribution has entropy $\frac{1}{2}(1+\log(2 \pi \sigma^2))$: if we increase $\sigma$ the entropy increases.
Because there is no maximal entropy for continuous densities over $R$ we must have additional constraints, e.g. the constraint that $\sigma$ is fixed and that $\mu$ is fixed. The fact that there is a given finite $\sigma^2$ and $\mu$ for me makes intuitively clear that there values nearer to $\mu$ must have higher probability. If you don't fix $\mu$ then you will get no unique solution.The Gaussian distribution for each real $\mu$ is a solution: this is some kind of "uniformness", all $\mu$ can be used for a solution.
Notice that it is crucial to fix $\sigma$, $\mu$ and to demand $p(x)>0 , \forall x \in R$. If you fix other values or change the form $R$ to another domain for the density funtion , e.g. $R^+$, you will get other solution: the exponential distribution, the truncated exponential distribution, the laplace distribution, the lognorma distribution (Theorems 3.3, 5.1, 5.2, 5.3)
Best Answer
Note that $Var(X)=E(X^{2})-(E(X))^{2}$ thus since $$E(X)=\frac{1}{n+1}\sum_{i=0}^{n}i=\frac{1}{n+1}\frac{(n+1)n}{2}=\frac{n}{2}$$ and $$E(X^{2})=\frac{1}{n+1}\sum_{i=0}^{n}i^{2}=\frac{1}{n+1}\frac{n(n+1)(2n+1)}{6}=\frac{n(2n+1)}{6}$$ Thus we have $$Var(X)=\frac{n(2n+1)}{6}-\left(\frac{n}{2}\right)^{2}=\frac{n^{2}+2n}{12}=\frac{(n+1)^{2}-1}{12}$$
Thus the derivation really relies on the fact that you know what sum of first n natural numbers is and sum of first n squared natural numbers. Hopefully this gives you the insight to do rest of problem.