This just says that a throw is less than $5$, that is, $1$, $2$, $3$, or $4$. We are asked for the probability of less than $3$, that is, $1$ or $2$, given that it is $1$, $2$, $3$, or $4$. We should not need theory to see that the probability is $2/4$.
Effectively, the condition "less than $5$" restricts the sample space to $4$ equally likely outcomes.
We can set it up and solve it as a formal conditional probability problem. Let $A$ be the event "less than $3$" and $B$ be the event "less than $5$." We want $P(A|B)$. Since $P(A|B)P(B)=P(A\cap B)$, we need only compute $P(B)$ and $P(A\cap B)$. Easily, the probability that a thrrow is less than $5$ is $4/6$. Also, $P(A\cap B)$ is just $P(A)$, which is $2/6$.
Remark: It is important not to be distracted by the irrelevant detail that there may have been a long string of $5$'s and/or $6$'s before the crucial throw. One could also do the computation by taking into account these irrelevant throws. More work, with result, when the smoke clears, $1/2$.
Computing the Probabilities
Using Inclusion-Exclusion to compute the number of ways to get all five faces ($1$-$5$) with $d-n$ dice, we get that the number of ways to have $n$ sixes with $d$ dice is
$$
\binom{d}{n}\left[\binom{5}{0}5^{d-n}-\binom{5}{1}4^{d-n}+\binom{5}{2}3^{d-n}-\binom{5}{3}2^{d-n}+\binom{5}{4}1^{d-n}-\binom{5}{5}0^{d-n}\right]
$$
Therefore, the probability of getting $2$ or more sixes, given that we have at least $1$ six is
$$
\frac{\displaystyle\sum_{n=2}^d\binom{d}{n}\left[\binom{5}{0}5^{d-n}-\binom{5}{1}4^{d-n}+\binom{5}{2}3^{d-n}-\binom{5}{3}2^{d-n}+\binom{5}{4}1^{d-n}-\binom{5}{5}0^{d-n}\right]}{\displaystyle\sum_{n=1}^d\binom{d}{n}\left[\binom{5}{0}5^{d-n}-\binom{5}{1}4^{d-n}+\binom{5}{2}3^{d-n}-\binom{5}{3}2^{d-n}+\binom{5}{4}1^{d-n}-\binom{5}{5}0^{d-n}\right]
}
$$
The results for different values of $d$:
$$
\begin{array}{c|c}
d&p&\text{approx}\\\hline
6&0&0.000000\\
7&\frac16&0.166667\\
8&\frac{17}{57}&0.298246\\
9&\frac{17}{42}&0.404762\\
\color{#C00000}{10}&\color{#C00000}{\frac{1606}{3261}}&\color{#C00000}{0.492487}\\
11&\frac{293}{518}&0.565637\\
12&\frac{18868}{30083}&0.627198
\end{array}
$$
Applying the Principle of Inclusion-Exclusion
Above, it is mentioned that we used inclusion-exclusion to compute the number of ways to get all five faces with $d-n$ dice.
To do this, we let $A_k$, for $k\in\{1,2,3,4,5\}$, be the arrangements without face $k$. Thus, the number of ways to arrange $d$ dice with some face missing, is
$$
\begin{align}
\left|\bigcup_k A_k\right|
&=\overbrace{\ \sum_k\left|A_k\right|\ }^{\binom{5}{1}4^d}-\overbrace{\sum_{k_1\lt k_2}\left|A_{k_1}\cap A_{k_2}\right|}^{\binom{5}{2}3^d}+\color{#9000B0}{\overbrace{\sum_{k_1\lt k_2\lt k_3}\left|A_{k_1}\cap A_{k_2}\cap A_{k_3}\right|}^{\binom{5}{3}2^d}}\\
&-\underbrace{\sum_{k_1\lt k_2\lt k_3\lt k_4}\left|A_{k_1}\cap A_{k_2}\cap A_{k_3}\cap A_{k_4}\right|}_{\binom{5}{4}1^d}+\underbrace{\sum_{k_1\lt k_2\lt k_3\lt k_4\lt k_5}\left|A_{k_1}\cap A_{k_2}\cap A_{k_3}\cap A_{k_4}\cap A_{k_5}\right|}_{\binom{5}{5}0^d}
\end{align}
$$
Let's explain how to compute the purple term, the total of the intersections of three of the $A_k$. The others will be similar.
$\binom{5}{3}$ is the number of ways to choose the $3$ faces to be excluded. Once those have been excluded, there are $2$ faces onto which to map the $d$ dice; that is, $2^d$ ways. Thus, the purple sum is
$$
\binom{5}{3}2^d
$$
Thus, to count the number of ways to get all five faces with $d$ dice, we subtract from the number of ways to arrange all $d$ dice:
$$
\binom{5}{0}5^d-\binom{5}{1}4^d+\binom{5}{2}3^d-\binom{5}{3}2^d+\binom{5}{4}1^d-\binom{5}{5}0^d
$$
which is used above.
Best Answer
There is duplication issue here.
Consider the following two cases:
($1$) You choose the first throw to be the same as $5$th throw, and the second throw happens to be the same as well.
($2$) You choose the second throw to be the same as $5$th throw, and the first throw happens to be the same as well.
They are essentially the same case but you have double counted these two cases.