Probability of toss of random Value X A random variable X has the probability distribution
value                1  ,       2        ,     3     ,         4
probability         0.5  ,     0.25      ,    0.125     ,   0.125
(For those of you who are interested, this is the geometric $p=0.5$ “killed” at $4$. $X$ is the number of times I toss a coin if I follow this rule: I’ll toss the coin till I get the first head, but I’ll stop after $4$ tosses even if I haven’t got a head by that time.)
Find $E(X)$; expected value and $SE(X)$; standard error of $X$?
$E(X)$ = Average = $(1*0.5)+(2*0.25)+(3*0.125+(4*0.125)=1.875$ ok
$SE(X)$=standard error of $X = SD= SQRt((1-1.875)^2 *0.5+(2-1.875)^2 *0.25+(3-1.875)^2 *0.125+(4-1.875)^2*0.125)=$
 A: You're given a random variable $X$ with probability mass function $p_X$ given by
$$
p_X(x)=P(X=x)=
\begin{cases}
0.5 &\quad\text{if }x=1 \\
0.25 &\quad\text{if }x=2 \\
0.125 &\quad\text{if }x=3 \\
0.125 &\quad\text{if }x=4
\end{cases}
$$
Now, to calculate the expected value you use the formula for a discrete random variable
$$
{\rm E}[X]=\sum_{x}x\cdot p_X(x).
$$
Note that this is in fact a weighted average of the values $1,2,3,4$ weighted by the probability of $X$ taking on that specific value. This is why your formula is incorrect, it fails to take the probabilities/weights into account.
Now, to calculate the standard error, you must first calculate the variance $\mathrm{Var}(X)$ of $X$. To do this we can use the calculation formula:
$$
\mathrm{Var}(X)={\rm E}[X^2]-{\rm E}[X]^2.
$$
We already know the last term and finding ${\rm E}[X^2]$ is an easy exercise using the law of the unconscious statistician stating that
$$
{\rm E}[X^2]=\sum_x x^2\cdot p_X(x).
$$
To find the standard error you simply take the square root of the variance.
Note that you can actually calculate the variance directly using the law of the unconscious statistician as
$$
{\rm Var}(X)={\rm E}[(X-\mu)^2]=\sum_x (x-\mu)^2\cdot p_X(x)
$$
where $\mu={\rm E}[X]$ is the expected value of $X$. Again you forgot to weight with the probabilities in your formula.
A: For a random variable that takes values $x_1,x_2,x_3,...,x_n$ with probabilities $p_1,p_2,p_3,...,p_n$, the expected value is:
$\mu=E[X]=x_1*p_1+x_2*p_2+x_3*p_3+...+x_n*p_n$
Your solution has $1*\frac{1}{4}+2*\frac{1}{4}+3*\frac{1}{4}+4*\frac{1}{4}$. This would be correct if each of the probabilities was $\frac{1}{4}$, but that is not the case here. 
The standard error is:
$\sigma=\sqrt{E[(X-\mu)^2]}=\sqrt{(x_1-\mu)^2*p_1+(x_2-\mu)^2*p_2+(x_3-\mu)^2*p_3+...+(x_n-\mu)^2*p_n}$
Your solution for the standard error does not include any probabilities (and the value of $\mu$ from the first part is incorrect). 
Does this help? 
