Ultimately show $R\cong S$ implies $R[x]\cong S[x]$. 
The question above is going to be used to ultimately show that $R\cong S$ implies $R[x]\cong S[x]$. I understand how these results imply that, but I am having trouble actually doing the proof parts for the question above. I have a hard time showing something is a homomorphism or injective/surjective when it is defined as above, the polynomial form really confuses me.
If it helps this is page 95 #21 in Thomas Hungerford's Abstract Algebra 3rd Edition.
 A: So you basically just need to carefully ask yourself: how is multiplication defined on $R[x]$? Well if $f(x)=\sum_{i=0}^n a_ix^i$ and $g(x)=\sum_{i=0}^mb_ix^i$, then $f(x)g(x)$ is given by $\sum_{k=0}^{n+m}c_kx^k$, where $c_k=\sum_{j=0}^k a_jb_{k-j}$. Therefore we have
$$\overline h(f(x)g(x))=\overline h\big(\sum_{k=0}^{n+m}c_kx^k\big)=\sum_{k=0}^{n+m}h(c_k)x^k.$$
Okay so what now? Well remind yourself what $c_k$ actually is, then use the fact that $h$ is a homomorphism to check that $h(c_k)=\sum_{j=0}^k h(a_j)h(b_{k-j})$.
Now separately use the definition of multiplication in $S[x]$ to calculate the product $\overline h(f(x))\overline h(g(x))$, i.e. the product $\big(\sum_{i=0}^n h(a_i)x^i\big)\big(\sum_{i=0}^m h(b_i)x^i\big)$, and notice the $k$-th coefficient of this new polynomial is exactly the coefficient $h(c_k)$ as above, and conclude $\overline h(f(x)g(x))=\overline h(f(x))\overline h(g(x))$.
For injectivity, remember we can check the kernel is zero. Suppose $f(x)=\sum_{i=0}^n a_ix^i$ is in the kernel of $\overline h$; that is, we have $\sum_{i=0}^n h(a_i)x^i=0$ in $S[x]$. Now conclude that each coefficient here is zero, and use the injectivity of $h$ to show that each $a_i$ is zero, i.e. that $f(x)=0$, and this concludes injectivity. 
Surjectivity should be straightforward once you've filled in the details for everything above, so I won't say anything about that.
