Write through and write back cache policies
When you want a general formula for a particular case, just derive it.

Let the main memory access time be $x$, and the cache access time be $y$. Let the hit ratio be $h$ and the fraction of requests that are reads be $p$.

Now,
$$\begin{align}\text{Average access time}&amp;=\text{Weighted average of read and write times}\\&amp;=\text{Fraction of reads}\times\text{Average read time}+\text{Fraction of writes}\times\text{Average write time}\tag1\end{align}$$

Let us now go through the read process. Firstly, there are two possibilites - a cache hit or a miss. If it is a hit, the cache access time is the total read time. If it is a miss, then the total read time is the sum of the time taken to access the cache and the time taken to access the memory. The fraction of hits is given by the hit ratio. Again, we take the weighted average.

$$\begin{align}\text{Average read time}&amp;=\text{Fraction of hits}\times\text{Cache access time}+\text{Fraction of misses}\times(\text{Cache access time}+\text{Memory access time})\\
&amp;=hy+(1-h)(x+y)\tag2\end{align}$$

Now, the writes. It is given that the write policy is the write through policy. This means that when a write is made, irrespective of a hit or a miss, memory is accessed as well. As the memory access time dominates over the cache access time, and the hit ratio for writes is not given, ignore the difference in the times of accessing just the memory and accessing the cache and the memory.

$$\begin{align}\text{Average write time}&amp;=\text{Main memory access time}=x\tag3\end{align}$$

Combining $(1),(2),(3)$, we get,
$$\begin{align}\text{Average access time}&amp;=p\left[hy+(1-h)(x+y)\right]+(1-p)x\end{align}$$

Here, $p=0.85,h=0.92,x=100\text{ns},y=10\text{ns}$. Substituting,
$$\begin{align}\text{Average access time}&amp;=0.85\left[0.92\times10+(1-0.92)(100)\right]+(1-0.85)\times100\\
&amp;=0.85\times17.2+15\\
&amp;=29.62\text{ns}\end{align}$$

Similarly, you can attempt the other cases mentioned in your question.