Just a remark?

If a number is divisible by $3$ and $8$ , then it is divisible by $24$ .

We will use that $a-b\mid a^n-b^n$ , where $a,b,n$ are three positive integers.

Divisible by $3$ :

$\begin{aligned} 13^n+3\times 5^{n-1}+8 & = (13^n-1)+3\times 5^{n-1}+9 \\ & = (13-1)\times a+3\times 5^{n-1}+9 \\ & = 3\times (4a+5^{n-1}+3) \end{aligned}$

So it is always divisible by $3$ .

Divisible by $8$ :

$\begin{aligned} S & =13^n+3\times 5^{n-1}+8 \\ & = 13^n-5^n+5^n+3\times 5^{n-1}+8 \\ & = (13^n-5^n)+8(5^{n-1}+1) \end{aligned}$

Since $(13-5)=8\mid 13^n-5^n$ , $S$ is always divisible by $8$ .

Therefore it is true.