First, some warm up:
Let's define our events at the start:
$\ \ \ D$ is the event that the student answers correctly.
$\ \ \ A$ is the event that the student guesses the answer.
$\ \ \ B$ is the event that the student copies the answer.
$\ \ \ C$ is the event that the student knows the answer.
Let's also write down what we know:
$$\textstyle
P(A)={1\over3},\quad P(B)={1\over 6},\quad P(D\mid B)={1\over 8}
.
$$
Also note
$$\textstyle
P(D\mid C)=1,
\quad P(D\mid A)={1\over 4},\quad P( C) =1-{1\over3}-{1\over6}={1\over2}.
$$
Now on to the problem proper:
You want to find $P(C\mid D)$.
$C$ and $D$ are not independent. We have to use the basic formula defining conditional probabilities:
$$\tag{1}
P(C\mid D) ={P(C\cap D)\over P(D)}.
$$
To find $P(C\cap D)$, we use the basic formula again (though it's usually called the multiplication principle when used this way):
$$
P(C\cap D) =P(C)P(D\mid C).
$$
We know $P(C)={1\over2}$ (as you calculated); and, if we're given that the student knows the answer, it follows that in this case that the probability that the student answers correctly is 1. Thus
$$\textstyle\tag{2}P(C\cap D) = {1\over2}\cdot 1={1\over 2}.$$
Now to find the term $P(D)$ in $(1)$, we first write
$$\tag{3}
P(D)=P(A\cap D)+P(B\cap D)+P(C\cap D)
$$
this is allowed since $A$, $B$, and $C$ are mutually exclusive events and one of the three must occur; as sets, $D$ can be written as the disjoint union $D= (A\cap D)\cup (B\cap D)\cup (C\cap D)$.
On to calculating the terms in $(3)$:
We have already calculated $P(C\cap D)$.
To find $P(A\cap D)$:
$$\tag{4}\textstyle
P(A\cap D)=P(A)P(D\mid A)={1\over3}\cdot{1\over4}={1\over12}.
$$
To find $P(B\cap D)$:
$$\tag{5}\textstyle
P(B\cap D)=P(B)P(D\mid B)={1\over6}\cdot{1\over8}={1\over48}.
$$
So, substituting the information from $(2)$, $(4)$ and $(5)$ into equation $(3)$, we have
$$\textstyle
P(D)= {1\over12}+{1\over48}+{1\over2} ={29\over 48}.
$$
Using this and $(1)$ and $(2)$ we finally obtain
$$
P(C\mid D) = {P(C\cap D)\over P(D)}={ 1/2\over 29/48}= {48\over 2\cdot 29}={24\over29}.
$$
Your reasoning for (2) is in error. Imagine a 3-sided die. The probability of $A=\{1,2\}$ occurring and $B=\{2,3\}$ occurring is equal and their union is a certain event.
You can get the answer quickly by elimination:
$$1=P(E_1 \cup E_2) < P(E_1)+P(E_2) = 2P(E_1) = 2P(E_2)$$
The strict less-than comes from being independent. So $P(E_1),P(E_2)$ are both larger than $1/2$
You can explicitly prove this, too
$$1=P(E_1 \cup E_2) = P(E_1)+P(E_2)-P(E_1)P(E_2) = 2P(E_1) - P(E_1)^2$$
$$\iff P(E_1)^2 - 2P(E_1)+1=(P(E_1)-1)^2=0$$
Best Answer
(This is a comment, not an answer, but I can't put an image in a comment.)
A better picture to have in mind for independence is something like: