1. Interpret the random variable

Let $X$ be the number of tosses required to get the first six. Then $X$ follows a geometric distribution with success probability $p=\frac16,$ and failure probability $q=1-p=\frac56.$

So, $P(X=n)=\left(\frac56\right)^{n-1}\left(\frac16\right), \quad n=1,2,3,\dots$

2. Write the required conditional probability

We need $P(X\ge 5\mid X>2).$ Using the formula for conditional probability, $P(X\ge 5\mid X>2)=\frac{P(X\ge 5\cap X>2)}{P(X>2)}.$

Since $X\ge 5$ automatically implies $X>2$, we have $X\ge 5\cap X>2 = X\ge 5.$ Therefore, $P(X\ge 5\mid X>2)=\frac{P(X\ge 5)}{P(X>2)}.$

3. Compute $P(X\ge 5)$

$X\ge 5$ means no six appears in the first 4 tosses. Thus, $P(X\ge 5)=\left(\frac56\right)^4=\frac{625}{1296}.$

4. Compute $P(X>2)$

$X>2$ means no six appears in the first 2 tosses. Thus, $P(X>2)=\left(\frac56\right)^2=\frac{25}{36}.$

5. Find the conditional probability