1. General term in the binomial expansion

For $\left(3^{1/4}+5^{1/8}\right)^{60},$ the general term is $T_{r+1}=\binom{60}{r}\left(3^{1/4}\right)^{60-r}\left(5^{1/8}\right)^r, \qquad r=0,1,2,\dots,60.$

So, $T_{r+1}=\binom{60}{r} \cdot 3^{\frac{60-r}{4}} \cdot 5^{\frac{r}{8}}.$

A term will be rational only when both exponents are integers:

• $\dfrac{60-r}{4}$ must be an integer,
• $\dfrac{r}{8}$ must be an integer.

2. Conditions for rationality

From $\frac{r}{8}\in \mathbb{Z},$ we need $r$ to be a multiple of $8$.

From $\frac{60-r}{4}\in \mathbb{Z},$ we need $60-r$ to be divisible by $4$. Since $60$ is divisible by $4$, this means $r$ must also be divisible by $4$.

Combining both, $r$ must be divisible by $8$.

Thus rational terms occur when $r=0,8,16,24,32,40,48,56.$

Number of rational terms: $8.$

3. Total number of terms

In the expansion of $(a+b)^{60}$, total terms are $60+1=61.$

So number of irrational terms is $n=61-8=53.$

4. Compute $n-1$

$n-1=53-1=52.$

Now check divisibility:

• $52$ is not divisible by $30$
• $52$ is not divisible by $8$
• $52$ is not divisible by $7$
• $52$ is divisible by $26$

Hence, the correct option is $\boxed{\text{D: }26}.$