Answer
$R(t)$ is continous on $R$.
Work Step by Step
$$R(t)=\frac{e^{\sin t}}{2+\cos\pi t}$$
1) Consider the numerator
We see that $\sin t$ is defined for $\forall t\in R$.
So, $e^{sin t}$ is also defined for $\forall t\in R$.
Therefore, the domain of $e^{\sin t}$ is $R$.
2) Consider the denominator
We know $2+ \cos\pi t$ is defined for $\forall t\in R$.
Therefore, the domain of $2+\cos\pi t$ is $R$.
3) Find the common domain of both the numerator and denominator.
Combine the results from 1) and 2), we conclude the common domain is $R$.
In other words, according to Theorem 7, both $e^{\sin t}$ and $2+\cos\pi t$ are continuous on $R$.
4) According to Theorem 4, $R(t)$ is also continous on $R$ except where $2+\cos\pi t=0$.
However, notice that $$-1\leq\cos\pi t\leq1$$$$1\leq2+\cos\pi t\leq3$$ for $\forall t\in R$
Which means, $2+\cos\pi t\ne0$ for $\forall t\in R$.
In conclusion, $R(t)$ is continous on $R$.
