We substitute $x=1$ into each expression to find $f(g(1))+g(f(1))$.
First, we find $g(1)=1-1=0$.
Next, we find $f(1)=\frac{3(1)^2+5(1)+8}{(1)^2-(1)+4}=\frac{3+5+8}{1-1+4}=\frac{16}{4}=4$.
Now, we substitute these values back into the expression: $f(g(1))+g(f(1))=f(0)+g(4)$.
To find $f(0)$, we substitute $x=0$ into $f(x)$: $f(0)=\frac{3(0)^2+5(0)+8}{(0)^2-(0)+4}=\frac{8}{4}=2$.
Finally, we find $g(4)=4-1=3$.
Putting it all together, we have $f(g(1))+g(f(1))=f(0)+g(4)=2+3=\boxed{5}$.
The answer is: 5
Below is a math problem. Provide a detailed, step-by-step solution.

### Problem:
Let $f(x)=\frac{3x^2+5x+8}{x^2-x+4}$ and $g(x)=x-1$. Find $f(g(x))+g(f(x))$, evaluated when $x=1$.

### Answer:
