Since you are really asking two questions, namely how to solve your specific inequalities and also how to go about proving inequalities in general, I will consider them separately and in that order.
Question 1
Deductions: Using your given information, there are a few observations that we can make that will be helpful later in the proofs of your inequalities (of course, I made these deductions while trying to prove the inequalities--I did not see how they related to your problem until I needed them).
$\underline{\text{Observation 1}}$:
Notice that
\begin{align}
z-1 &= x+y-1\tag{$z=x+y$}\\[0.5em]
&= u+y-1\tag{$x=u$}\\[0.5em]
&< u.\tag{$y<1$; thus, $y-1<0$}
\end{align}
Since we are also given that $u<1$, we effectively have that
$$
z-1<u<1.\tag{$\dagger$}
$$
$\underline{\text{Observation 2}}$: Notice that
$$
\frac{z}{2} = \frac{x+y}{2} = \frac{x}{2}+\frac{y}{2}<1;\qquad\text{(since $x<1$ and $y<1$)}
$$
also,
$$
\frac{z}{2}=\frac{u+y}{2}=\frac{u}{2}+\frac{y}{2}>\frac{u}{2}+\frac{u}{2}=u.
$$
Since $\frac{z}{2}<1$ and $\frac{z}{2}>u$, we effectively have that
$$
u < \frac{z}{2} < 1.\tag{$\dagger\dagger$}
$$
Givens: Using your initial givens and the observations $\dagger$ and $\dagger\dagger$ we just established, let's list out all of the givens we currently have to work with.
\begin{align}{}
0<x<y<1\tag{1}\\[0.5em]
z=x+y\tag{2}\\[0.5em]
x=u\tag{3}\\[0.5em]
y=z-u\tag{4}\\[0.5em]
0<u<z-u<1\tag{5}\\[0.5em]
z-1<u<1\tag{6}\\[0.5em]
u<\frac{z}{2}<1\tag{7}
\end{align}
Now proving your inequalities will actually be fairly straightforward.
Inequality 1 ($0<z<1\to 0<u<\frac{z}{2}$): Start with the left-hand side and work towards the right-hand side using the information you are given (the approach will be the same for the second inequality):
\begin{align}
0<z<1 &\to 0<x+y<1\tag{since $z=x+y$}\\[0.5em]
&\to 0<2x<x+y<1\tag{since $x<y$ and $x>0$}\\[0.5em]
&\to 0<2u<z<1\tag{since $x=u$ and $z=x+y$}\\[0.5em]
&\to 0<u<\frac{z}{2}<\frac{1}{2}\tag{divide throughout by $2$}\\[0.5em]
&\to 0<u<\frac{z}{2}.\tag{desired expression}
\end{align}
Thus, we see that $0<z<1\to 0<u<\frac{z}{2}$, concluding the proof of the first inequality.
Inequality 2 ($1<z<2\to z-1<u<\frac{z}{2}$): Starting with the left-hand side,
\begin{align}
1<z<2 &\to 0<z-1<1\tag{subtract $1$ throughout}\\[0.5em]
&\to 0<z-1<u<1\tag{since $z-1<u<1$}\\[0.5em]
&\to 0<z-1<u<\frac{z}{2}<1\tag{since $u<\frac{z}{2}<1$}\\[0.5em]
&\to z-1<u<\frac{z}{2}\tag{desired expression}
\end{align}
we reach the right-hand side. Thus, we see that $1<z<2\to z-1<u<\frac{z}{2}$, concluding the proof of the second inequality.
Question 2
In regards to proving inequalities in general, you need to realize it truly is an art form, and, as with most forms of art, it takes time and practice to get better. Numerous books have been written just on inequalities and how to strategically attack them. There is such a great range of difficulty when it comes to proving inequalities. Here is a sample of inequalities that differ greatly in difficulty (links are to answers I have provided to such inequality problems on MSE):
- Easy
- Average
- Hard
- Very hard
The more problems you encounter and solve, the better and better you will get at solving them. I will conclude my answer with an example.
Example: Compare $6^{1/4}$ and $4^{1/3}$ without using a calculator. Is one greater than the other?
Answer. Observe that
$$
6^{\frac{1}{4}}=6^{\frac{3}{12}}=(6^3)^{\frac{1}{12}}=216^{\frac{1}{12}}
$$
and
$$
4^{\frac{1}{3}}=4^{\frac{4}{12}}=(4^4)^{\frac{1}{12}}=256^{\frac{1}{12}}.
$$
Since $256$ and $216$ are both being raised to the same power, and $256>216$, we can clearly see that $4^{1/3} > 6^{1/4}$ since
$$
4^{\frac{1}{3}} = 256^{\frac{1}{12}} > 216^{\frac{1}{12}} = 6^{\frac{1}{4}}.
$$
Knowing that it helps to make such manipulations comes with time and practice. Notice that it would otherwise be quite difficult to compare the values in the example as $4^{1/3}\approx 1.5874$ and $6^{1/4}\approx 1.5651$.
