It looks fine to me. What you’re using is the fact that if the random variables $X$ and $Y$ are uniformly distributed in $\{0,1,\dots,n-1\}$, then so is the reduced sum $Z=(X+Y)\bmod n$, where $\bmod$ here denotes the binary operation. (In your case $n$ is of course $10$.)
To see this, you can observe that $X+Y$ itself takes values in the set $\{0,1,\dots,2n-2\}$, with the following probabilities:
$$\mathbb{P}[X+Y=k]=\begin{cases}
\frac{k+1}{n^2},&k\le n-1\\\\\\
\frac{2n-1-k}{n^2},&k\ge n-1\;.
\end{cases}$$
(Just count the number of ways that each sum can occur.)
But $Z=(X+Y)\bmod n=k$ iff $X+Y=k$ or $X+Y=k+n$, so
$$\mathbb{P}[Z=k]=\frac1{n^2}\bigg((k+1)+\big(2n-1-(k+n)\big)\bigg)=\frac1{n}\;,$$
i.e., $Z$ is uniformly distributed in $\{0,1,\dots,n-1\}$. By induction the same is true for any finite sum reduced modulo $n$.
$$\color{red}{416/3125}=0.13312.
$$
The last digit must be $2$ or $4$, this happens with probability $2/5$.
The sum of the four other digits must be $\pm1\pmod{3}$, according to the last digit being $2$ or $4$. Since both events have the same probability, the answer is $2/5$ times the probability that the sum $s$ of four digits is $1\pmod{3}$, that is, $s=-2$ or $s=+1$ or $s=+4$.
$s=+4$ corresponds to $+1,+1,+1,+1$, with probability $2^4/5^4$.
$s=+1$ corresponds to $0,0,0,+1$, or $0,+1,+1,-1$ in whatever order. In the first case, one must place the $+1$, thus $4$ cases, with probability $2/5^4$ each. In the second case, one must place the $0$ and the $-1$, thus $12$ cases, with probability $2^3/5^4$ each.
$s=-2$ corresponds to $+1,-1,-1,-1$, thus $4$ cases, with probability $2^4/5^4$ each, or to $0,0,-1,-1$, thus $6$ cases, with probability $2^2/5^4$ each.
Summing up, the answer is $(2/5)\cdot(2^4+4\cdot2+12\cdot2^3+4\cdot2^4+6\cdot2^2)/5^4$.
Best Answer
Let's split the $n$-digit numbers without consecutive identical digits that end in a given digit, with leading zeros allowed, into $a_n$ numbers with identical first and last digits and $b_n$ numbers with different first and last digits. Then we have the recurrences $a_{n+1}=b_n$ and $b_{n+1}=9a_n+8b_n$. Eliminating $b_n$ yields $a_{n+2}=9a_n+8a_{n+1}$. With $a_1=1$ and $a_2=0$, this yields $a_3=9$, $a_4=72$, $a_5=657$, $a_6=5904$ and thus also $b_5=5904$. We want the $a_5$ admissible numbers ending in $5$, and the $8/9$ of the $b_5$ admissible numbers ending in $5$ that don't begin with $0$, and the $b_5$ admissible numbers ending in $0$. So that's
$$ a_5+\frac89b_5+b_5=657+\frac{17}9\cdot5904=11809\;. $$
This we have to divide by the total number $9\cdot10^4$ of $5$-digit numbers without leading zeros to arrive at your computed probability.