Is this the correct solution? Determine the coordinates of the vector $U=(4,5,-3)\;\text{of}\; R^3$ with respect to base ${(1,0,0), (0,1,0), (0,0, 1)}$
$$x(1,0,0) + y (0,1,0) + z (0,0,1) = (4,5, -3)$$
$$(x, 0,0) + (0, y, 0) + (0,0,z) = (4,5, -3)$$
$$x +0 +0 = 4 \Longrightarrow x = 4$$
$$0 +0 + y = 5 \Longrightarrow y = 5$$
$$0 +0 + z = -3 \Longrightarrow z = -3$$
Thus, the coordinates are $$(4,5, -3).$$
$$$$ $$$$And the base $\{(1,1,1),(1,2,0),(3,1,0)\}$, just find the values ​​of $x$, $y$ and $z$ such that $$x(1,1,1)+y(1,2,0)+z(3,1,0)=(4,5,-3)??$$
$$(x,x,x)+(y,2y,0)+(3z,z,0)=(4,5,-3)$$$\begin{cases}x+y+3z=4&\\ x+2y+z=5&\\ x+0+0=-3&\end{cases}$$$x=-3$$$$y=\frac{53}{5}$$and$$z=-\frac{6}{5}$$
Is this correct?

Yes, note that definition of representation of a vector $\vec v$ with respect to a basis $B=\left\langle {\vec{\beta}_{1},...,\vec{\beta}_{n}} \right\rangle $ of a $n$-dimensional space $V$ is
$$Rep_{B}(\vec v)=\left( {\begin{array}{*{20}{c}}
  {c_{1}} \\ 
  { \vdots } \\ 
  { c_{n} }
\end{array}} \right)$$
where $c_{1}, ..., c_{n}$ are coefficients such that $\vec v=c_{1}\vec{\beta}_{1}+...+c_{n}\vec{\beta}_{n}$. With your basis 
$$B=\left\langle  \left( {\begin{array}{*{20}{c}}
  { 1 } \\ 
  { 1 } \\ 
  { 1 }
\end{array}} \right),
\left( {\begin{array}{*{20}{c}}
  { 1 } \\ 
  { 2 } \\ 
  { 0 }
\end{array}} \right),
\left( {\begin{array}{*{20}{c}}
  { 3 } \\ 
  { 1 } \\ 
  { 0 }
\end{array}} \right)  
 \right\rangle$$
and 
$$\vec v = \left( {\begin{array}{*{20}{c}}
  { 4 } \\ 
  { 5 } \\ 
  { -3 }
\end{array}} \right)$$
then you have to find $c_1,c_2,c_3$ such that
$$
c_1 \left( {\begin{array}{*{20}{c}}
  { 1 } \\ 
  { 1 } \\ 
  { 1 }
\end{array}} \right) +
c_2 \left( {\begin{array}{*{20}{c}}
  { 1 } \\ 
  { 2 } \\ 
  { 0 }
\end{array}} \right) +
c_3 \left( {\begin{array}{*{20}{c}}
  { 3 } \\ 
  { 1 } \\ 
  { 0 }
\end{array}} \right) =
 \left( {\begin{array}{*{20}{c}}
  { 4 } \\ 
  { 5 } \\ 
  { -3 }
\end{array}} \right)
 $$
so solve this system and then your answer will be
$$Rep_{B} (
\left({\begin{array}{*{20}{c}}
  { 4 } \\ 
  { 5 } \\ 
  { -3 }
\end{array}} \right))=
\left({\begin{array}{*{20}{c}}
  { c_1 } \\ 
  { c_2 } \\ 
  { c_3 }
\end{array}} \right).$$
