If $U = (1, 2)$, $V = (3, -4)$, is the answer to $2U + \frac{1}{2}V$ the vector $(3.5, 2)$? Check my answer. If $U = (1, 2)$, $V = (3, -4)$, is the answer to $2U + \frac{1}{2}V$ the vector $(3.5, 2)$?
I did the following:
\begin{align*}
U : ( ( 2   * 1   ),  ( 2   *  2 ) ) &= ( 2  ,  4 )\\
V : ( ( 0.5 * 3   ),  ( 0.5 * -4 ) ) &= ( 1.5, -2 )\\
( ( 2   + 1.5 ) + ( 4   + -2 ) ) &= ( 3.5,  2 )
\end{align*}
 A: Your final answer is correct, but some of your intermediate steps could be considered wrong (or at least unclear).
First of all, you write 
$$U : (2 * 1, 2 * 2) = (2, 4).$$ 
It's not clear at all what you mean by this. What you have done is calculated $2U$. A much better way of writing this is 
$$2U = (2 * 1, 2 * 2) = (2, 4).$$
Likewise, you should write 
$$\frac{1}{2}V = (0.5 * 3, 0.5 * (-4)) = (1.5, -2).$$ 
As for your final line, you haven't indicated what you are doing. You are adding $2U$ and $\frac{1}{2}V$ so you should write
$$2U + \frac{1}{2}V = (2 + 1.5, 4 + (-2)) = (3.5, 2).$$
In summary, I think it would be better to format your answer as follows:

\begin{align*}
2U &= (2 * 1, 2 * 2) = (2, 4)\\
&\\
\frac{1}{2}V &= (0.5 * 3, 0.5 * (-4)) = (1.5, -2)\\
&\\
2U + \frac{1}{2}V &= (2 + 1.5, 4 + (-2)) = (3.5, 2).
\end{align*}

Alternatively, you could do all of the calculations at the same time:

$$2U + \frac{1}{2}V = (2 * 1, 2 * 2) + (0.5 * 3, 0.5 * (-4)) = (2, 4) + (1.5, -2) = (3.5, 2).$$


These comments are less important than the ones made above, but I think still worth pointing out. They are more about style than content.
In is better to use $\cdot$ or $\times$ to denote multiplication rather than $*$.
You shouldn't swap between fractions and decimals unless you have a good reason. As the question involved a fraction, use fractions the whole way through.
With these points in mind, I would present the calculation as follows:

\begin{align*}
2U + \frac{1}{2}V &= 2(1, 2) + \frac{1}{2}(3, -4)\\ 
&= (2\times 1, 2\times 2) + \left(\frac{1}{2}\times 3, \frac{1}{2}\times (-4)\right)\\ 
&= (2, 4) + \left(\frac{3}{2}, -2\right)\\ 
&= \left(\frac{5}{2}, 2\right).
\end{align*}

