Proof by induction that $1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}$ I am trying to understand how to do proof by induction for inequalities. The step that I don't fully understand is making an assumption that n=k+1. For equations it is simple. For example:

Prove that 1+2+3+...+n = $ \frac {n(n+1)}{2} $ is valid for $ i \ge 1 $ 

1) Base case for n=1. (...)
2) Assume that equation is true for n=k. (...)
3) Proof for n=k+1.
1+2+3+...+k+(k+1) = $ \frac {(k+1)[(k+1)+1]}{2} $
So we add +1 to k on both sides.
For inequalities:
Prove that $ 4n \lt 2^n $ is valid for $ n \ge 5 $
1) Base case. (...)
2) n=k assumption. (...)
3) n=k+1 proof.  
How I would do this(add 1 to k on both sides):
$ 4(k+1) < 2^{k+1} $
How it's done in tutorials:
$ 4(k+1) < 2^k + 4 $
$ 4k+4 < 2^k + 4 $
Why they add extra 4 to RHS instead of increasing k by 1?

For the first part, you haven't done anything, you just rewrote everything. Most importantly, you didn't prove the statement. Induction works by first showing that $P(1)$ holds then assuming $P(n)$ and proving $P(n+1)$. From there, it follows that the $n$ such that $P(n)$ is true must be all of $\mathbf{N}$. 
In our case $P(n)$ is the statement that $1+2+\dots+n=\frac{n(n+1)}{2}$. If we look at $n=1$, we see that $\frac{n(n+1)}{2}=\frac{1\cdot 2}{2}=1$. So that $P(1)$ holds. 
Now, we let $n$ be such that $P(n)$ holds and prove that $P(n+1)$ must also be true. Let's consider $1+2+\dots+n+(n+1).$ Recall that we know $P(n)$ is true, so, we can substitute $\frac{n(n+1)}{2}$ in for $1+\dots+n$. Therefore, $$1+2+\dots+n+(n+1)=\frac{n(n+1)}{2}+n+1=\frac{n^2+n+2n+2}{2}=\frac{(n+1)((n+1)+1)}{2}.$$ Notice that the far right hand side is exactly what we wanted to show, that is, that $P(n+1)$ holds.
So, we've shown that the set of $n$ such that $P(n)$ holds is all of $\mathbf{N}$, that is, all $n\geq 1$. 

Now, for the second question. We let $Q(n)$ be the statement $4n<2^n$ for $n\geq 5$ (I'm assuming yours is a typo, clearly $8=4\cdot 2< 2^2=4$ is not true.) 
So, let's do the base case: $2^5=32$ and $4\cdot 5=20$ so clearly we have $4\cdot 5< 2^5$. Thus, $Q(5)$ is true. 
Now, let $n$ be such that $Q(n)$ holds. We show that $Q(n+1)$ holds too. Observe that $4(n+1)=4n+4<2^n+4$ by the assumption. However, let's recall that $4=2^2<2^n$ for $n> 2$. Therefore, $$4(n+1)<2^n+2^n=2(2^n)=2^{n+1}.$$ 
Notice that we've only shown that $Q(n)$ implies $Q(n+1)$ when $n>2$. But, since $5>2$, it follows that $Q(n)$ is true for all $n\geq 5$. Which is exactly what we wanted to show. 
Hopefully this helps.
