How do we evaluate $\sum _{n=0}$ $\frac{8^n}{(3n)!} $?
We know that $e^z$= $\sum _{n=0}$ $\frac{z^n}{(n)!} $
So can we say that $\sum _{n=0}$ $\frac{8^n}{(3n)!} $ = $\frac{e^8}{3}$ ? Obviously the denominator is wrong. (How?)
Where do we put the $3$ then?
Also, $\sum _{n=0}$ $\frac{27^n}{(3n+1)!} $ $=?$ $\frac{e^{27}}{3n+1}$
You can evaluate this using a differential equation, but it's not very elegant. Define $$f(x)=\sum_{n=0}^\infty\frac{x^{3n}}{(3n)!}$$ and observe that this satisfies $$f''+f'+f=e^x,\\ f(0)=1,\quad f'(0)=0$$ Assuming you know how to solve a second-order linear differential equation, I'll spare you the details in showing that the general solution to $f''+f'+f=e^x$ is $$f(x)=\frac13e^x+e^{-x/2}(A\cos\tfrac{\sqrt3}2x+B\sin\tfrac{\sqrt3}2x)$$ You can use the initial data to show $A=\frac23$ and $B=0$. Hence $$\sum_{n=0}^\infty\frac{8^n}{(3n)!}=f(2)=\frac13e^2+\frac2{3e}\cos\sqrt3$$
Let's solve the more general case of $\sum\limits_{n=0}^\infty\frac{z^{an+b}}{(an+b)!}, b<a$. Through repeated differentiation, it can be seen that it satisfies the differential equation $$y=y^{(a)}, y^{(b)}(0)=1, y^{(k)}(0)=0, k<a, k\neq b$$ If we try the first sum of yours ($a=3,b=0$), we have $$y=y^{(3)},y(0)=1, y'(0)=0, y''(0)=0$$ which implies that $$\sum_{n=0}^\infty\frac{x^{3n}}{(3n)!}=\frac13\,{{\rm e}^{x}}+\frac23\,{{\rm e}^{-\frac12\,x}}\cos \left( \frac12\,\sqrt {3}x \right) $$ Plugging in $x=2$ gets us $$\sum_{n=0}^\infty\frac{2^{3n}}{(3n)!}=\sum_{n=0}^\infty\frac{8^{n}}{(3n)!}=\frac13e^2+\frac2{3e}\cos(\sqrt3)$$
As for your second question we can solve the differential equation for $a=3,b=1$ to get (it turns out to be the second derivative of the first function) $$\sum_{n=0}^\infty\frac{x^{3n+1}}{(3n+1)!}=\frac13\,{{\rm e}^{x}}-\frac13\,{{\rm e}^{-\frac12\,x}}\cos \left( \frac12\,\sqrt {3}x \right)+\frac1{\sqrt3}\,{{\rm e}^{-\frac12\,x}}\sin \left( \frac12\,\sqrt {3}x \right)$$ Plugging in $x=3$ and dividing by 3 we get $$\frac13\sum_{n=0}^\infty\frac{3^{3n+1}}{(3n+1)!}=\sum_{n=0}^\infty\frac{27^{n}}{(3n+1)!}=\frac19\,{{\rm e}^{3}}-\frac19\,{{\rm e}^{-\frac32}}\cos \left( \frac32\,\sqrt {3} \right)+\frac1{3\sqrt3}\,{{\rm e}^{-\frac32}}\sin \left( \frac32\,\sqrt {3} \right)$$
