Your comment answers both parts of the question. Since $a$ is a non-zerodivisor of $R$, it does not lie in any of the minimal primes of $R$. Therefore any chain of primes in $R$ that all contain $(a)$ has no minimal prime in the chain and can therefore be extended to a larger chain of primes in $R$, namely, by including a prime properly contained in the smallest prime in the chain. So $\dim R/(a) < \dim R$.

Since $a$ is a non-zerodivisor of $R$, it does not lie in any of the minimal primes of $R$. Therefore any chain of primes in $R$ that all contain $(a)$ has no minimal prime in the chain and can therefore be extended to a larger chain of primes in $R$, namely, by including a prime properly contained in the smallest prime in the chain. So $\dim R/(a) < \dim R$.

Your comment answers both parts of the question. Since $a$ is a non-zerodivisor of $R$, it does not lie in any of the minimal primes of $R$. Therefore any chain of primes in $R$ that all contain $(a)$ has no minimal prime in the chain and can therefore be extended to a larger chain of primes in $R$, namely, by including a minimal prime contained in the smallest prime in the chain. So $\dim R/(a) < \dim R$.

Your comment answers both parts of the question. Since $a$ is a non-zerodivisor of $R$, it does not lie in any of the minimal primes of $R$. Therefore any chain of primes in $R$ that all contain $(a)$ has no minimal prime in the chain and can therefore be extended to a larger chain of primes in $R$, namely, by including a prime properly contained in the smallest prime in the chain. So $\dim R/(a) < \dim R$.