Expected number of die rolls to get 6 given that all rolls are even.

Call $E_k$ the event "the first $k-1$ throws gave even results", and call $E$ the event "the first $6$ exists and is preceded by even numbers only".

I'll skip the clever answer from Mike Earnest's link and try to give an understandable scholar way.

The definition of a conditional expectation is $$\Bbb E(X|E)=\sum_{k=1}^\infty k\cdot\Bbb P(X=k\ |\ E)$$

Now $$\Bbb P(X=k\ |\ E)=\frac{\Bbb P((X=k)\cap E)}{\Bbb P(E)}$$ Note that the event $(X=k)\cap E$ is equal to the event $(X=k)\cap E_k$. $$\Bbb P((X=k)\cap E_k)=\left(\frac26\right)^{k-1}\cdot \frac16$$ as you have to independently get $2$s or $4$s at each of the $k-1$ first throws, and then get a $6$.

On the other hand, \begin{align*}\Bbb P(E)&=\sum_{k=1}^\infty \Bbb P(E\cap(X=k))\\&=\sum_{k=1}^\infty \left(\frac26\right)^{k-1}\cdot \frac16\\&=\frac16\cdot \frac{1}{1-\frac26}\\&=\frac14\end{align*}

Putting it all together: \begin{align*}\Bbb E(X|E)&=\sum_{k=1}^\infty k\cdot4\cdot \left(\frac26\right)^{k-1}\cdot \frac16\\&= \frac46 f'\left(\frac26\right)\\\end{align*} where $$f(x)=\sum_{k=0}^\infty x^k=\frac 1{1-x}\qquad\text{and}\qquad f'(x)=\sum_{k=1}^\infty k\cdot x^{k-1}=\frac 1{(1-x)^2}$$

Finally $$\Bbb E(X|E)=\frac23\cdot \frac1{\left(1-\frac26\right)^2}=\frac32.$$

We are throwing a die until we obtain neither $2$ nor $4$. The expected number of throws then is ${3\over2}$, by the formulas for a geometric random variable. If the closing throw is $1$, $3$, or $5$, we don't count the experiment and start afresh. If the closing throw is a six it's a success.