1. Given data

We have a twice differentiable function $f$ on $(1,6)$ such that:

$f(2)=8, \quad f'(2)=5, \quad f'(x)\ge 1, \quad f''(x)\ge 4 \text{ for all } x\in(1,6).$

We must test the options.

2. Use the condition on $f''(x)$

Since

$f''(x)\ge 4,$

integrate from $2$ to $x$:

$f'(x)-f'(2)=\int_2^x f''(t)\,dt \ge \int_2^x 4\,dt = 4(x-2).$

Using $f'(2)=5$,

$f'(x)\ge 5+4(x-2)=4x-3.$

In particular, at $x=5$,

$f'(5)\ge 4(5)-3=17.$

3. Now estimate $f(5)$

Using

$f(5)-f(2)=\int_2^5 f'(x)\,dx,$

and the bound $f'(x)\ge 4x-3$,

$f(5)-8 \ge \int_2^5 (4x-3)\,dx.$

Compute the integral:

$\int (4x-3)dx = 2x^2-3x.$

So,

$\int_2^5 (4x-3)dx = (2\cdot 25-15)-(2\cdot 4-6)=35-2=33.$

Hence,

$f(5)\ge 8+33=41.$

4. Check each option

Option A: $f(5)\le 10$

But we found

$f(5)\ge 41,$

so this is false.

Option B: $f(5)+f'(5)\ge 28$

We have

$f(5)\ge 41, \quad f'(5)\ge 17.$

Therefore,

$f(5)+f'(5)\ge 41+17=58\ge 28.$

So B is true.

Option C: $f(5)+f'(5)\le 26$

But

$f(5)+f'(5)\ge 58,$

so this is false.

Option D: $f'(5)+f''(5)\le 20$

We only know

$f'(5)\ge 17, \quad f''(5)\ge 4,$

hence

$f'(5)+f''(5)\ge 21.$

So it cannot be $\le 20$. Thus D is false.

5. Conclusion

The only correct option is:

$\boxed{\text{B}}$