Question on first and second price sealed bid auction

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I’m studying on auction. But I couldn’t understand some questions about this topic. And I cannot solve such type questions. For example,

I know that a set of symmetric and risk neutral bidders participate in a first-price sealed bid auction. For the number of bidders $n\ge 0$,

F(v) = the distribution function from which bidders’ valuations (independently drawn and it has uniform distribution $v\in [0,1]$, similarly f(v) is density function. $b(v)$ is strategies in auction and calculated by

$$b(v)=\int_0^v x \frac{(n-1)F(x)^{n-2}f(x)}{F(v)^{n-1}} dx$$

Firstly I try to show that $b(v) is increasing in v. Secondly, how can I show that b(v) is equilibrium strategy.

As for second price sealed bid, I try to show there exist an equilibrium for all bidders to bid their true valuations. And finally I want to compare the expected selling prices in the two auctions for conditional on a fixed highest valuation on $v_n$ by arranging bidders’ valuations from highest to lowest $v_1\ge … \ge v_n$. Well,how can I decide which auctions would a risk averse sellers prefer?

Question is this. And I cannot produce any proper solution for such type questions.

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now, I mention about my solution trials;

I can only show that the bidding function b(v) is increasing in v.

For that I rewrite the bidding function as follows:

$$b(v)=\frac{1}{F(v)^{n-1}}\int_0^v x dF^{n-1}(x)dx$$

Since each bidders value is uniformly distributed on [0,1], I can say F(v)=v and f(v)=1, so if there are n bidders then each employs the bidding function

$$b(v)={1\over v^{n-1}}\int^v_0 xdx^{n-1}={1\over v^{n-1}}\int^v_0 x(n-1)x^{n-2}dx$$ $$={n-1\over v^{n-1}}\int^v_0 x^{n-1}dx={1\over v^{n-1}}{1\over n} v^n = v-(v/N)=((n-1)v)/n$$

So as v increases, the bidding function will increase.

However, after this part of the question, I cannot solve. I also read some materials (like that https://dvikan.no/ntnu-studentserver/kompendier/14%20jehle%20reny.pdf ) but I could not. Please help me to solve these types questions. Thanks a lot.