The first sentence means that the number of dice rolled is a random variable $N$ such that $\Bbb P(N=n)$, the probability of rolling $n$ dice, is $\frac1{2^n}$. Thus, the probability of rolling one die is $\frac12$, the probability of rolling two dice is $\frac1{2^2}=\frac14$, and so on. Since $\sum_{n\ge 1}\frac1{2^n}=1$, this makes sense.
(a) If $S_n=4$, we can’t possibly have more then $4$ dice, so if $n$ is even, $n$ must (as you said) be $2$ or $4$, and the favorable rolls are exactly the ones that you specified. We want $\Bbb P(S_N=4\mid N\text{ is even})$, which by definition is
$\Bbb P(S_N=4\mid N\text{ is even})=\frac{\Bbb P(S_N=4\text{ and }N\text{ is even})}{\Bbb P(N\text{ is even})}\;,$
so we should try to calculate the probabilities on the righthand side.
$\begin{align*} \Bbb P(N\text{ is even})&=\sum_{k\ge 1}\Bbb P(N=2k)=\sum_{k\ge 1}\frac1{2^{2k}}\\ &=\sum_{k\ge 1}\frac1{4^k}=\frac{\frac14}{1-\frac14}\\ &=\frac13\;. \end{align*}$
If $X_k$ is the random variable representing the number on the $k$-th die,
$\begin{align*} \Bbb P(S_N=4\text{ and }N\text{ is even})&=\Bbb P(S_N=4\text{ and }N=2)+\Bbb P(S_N=4\text{ and }N=4)\\ &=\Bbb P\big(N=2\text{ and }X_1+X_2=4\big)\\ &\qquad+\Bbb P(N=4\text{ and }X_1=X_2=X_3=X_4=1)\\ &=\frac1{2^2}\cdot\left(3\left(\frac16\right)^2\right)+\frac1{2^4}\cdot\left(\frac16\right)^4\\ &=\frac14\cdot\frac3{36}+\frac1{16}\cdot\frac1{1296}\\ &=\frac1{48}+\frac1{20736}\\ &=\frac{433}{20736}\;. \end{align*}$
Thus, $\Bbb P(S_N=4\mid N\text{ is even})=\frac{\frac{433}{20736}}{\frac13}=\frac{433}{6912}\approx0.0626\;.$
This should get you started; I’ll add a bit more later.
Added: For (b) we want $\Bbb P(N=2\mid S_N=3)$. If $S_N=3$, there are three possibilities: $N=1$ and $X_1=3$; $N=2$ and either $X_1=1$ and $X_2=2$, or $X_1=2$ and $X_2=1$; or $N=3$ and $X_1=X_2=X_3=1$. Thus,
$\begin{align*} \Bbb P(S_N=3)&=\Bbb P(N=1\text{ and }S_N=3)+\Bbb P(N=2\text{ and }S_N=3)+\Bbb P(N=3\text{ and }S_N=3)\\ &=\frac12\cdot\frac16+\frac1{2^2}\cdot\frac2{36}+\frac1{2^3}\left(\frac16\right)^3\\ &=\frac{169}{1728}\;. \end{align*}$
The term $\dfrac1{2^2}\cdot\dfrac2{36}=\dfrac1{72}$ corresponds to the cases in which $N=2$: it’s the probability that $S_N=3$ and $N=2$. It’s share of the total probability of the event $S_N=3$ is therefore
$\Bbb P(N=2\mid S_N=3)=\frac{\frac1{72}}{\frac{169}{1728}}=\frac{1728}{72\cdot169}=\frac{24}{169}\approx0.1420\;.$
I’ll leave the other two to you, but feel free to ask questions if you get completely stuck.