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.  In general, if we're given the coordinates of a vector in $\mathbb R^n$, those are the coordinates with respect to the standard base of $\mathbb R^n$.
