How does derandomization of 3SAT work via conditional expectations?
If you choose an assignment at random, each particular clause is satisfied with probability $7/8$. Using linearity of expectation, this shows that if you choose an assignment at random, then the number of satisfied clauses is at least $(7/8)n$. The idea is to have an indicator variable $X_i$ for clause $i$, which states that the clause is satisfied. Since $\mathbb{E}[X_i] = \Pr[X_i] = 7/8$, by linearity of expectation we have $\mathbb{E}[\sum_{i=1}^m X_i] = (7/8)m$. In particular, some assignment satisfies at least $(7/8)m$ clauses.

Let now $X$ be the number of unsatisfied clauses under a random assignment. Then $\mathbb{E}[X] = (1/8)m$, and so Markov's inequality shows that
$$ \Pr[X \geq (C/8)m] \leq 1/C. $$
By choosing an appropriate $C$ you get your first bullet (try it out!).

Regarding the method of conditional expectations, there are many online resources, for example this one (the first hit on a popular search engine). There are many others.

Finally, if $\mathcal{D}$ is a 3-wise independent sample space – this just means that if $(v_1,\ldots,v_n) \sim \mathcal{D}$, then for any $i&lt;j&lt;k$, the joint distribution of $(v_i,v_j,v_k)$ is uniform over all eight possibilities – then it is still the case that each clause is satisfied with probability exactly $7/8$ (why?). Therefore if we choose a random assignment according to $\mathcal{D}$, the expected number of satisfied assignments is exactly $(7/8)m$. In particular, some assignment in the support of $\mathcal{D}$ – that is, an assignment which is generated by $\mathcal{D}$ with some probability – satisfies at least $(7/8)m$ clauses.

Why is this helpful? There are 3-wise independent samples spaces of polynomial size. We can derandomize the random 3SAT algorithm by taking such a sample space and trying out all assignments in its support. One of them will satisfy at least $(7/8)m$ clauses.