1. Since each branch is differentiable on its own interval, the only point to check is the junction at $x=1$.

2. Continuity at $x=1$

For $0<x<1$, $f(x)=3x^2+k\sqrt{x+1}$ So, $\lim_{x\to 1^-} f(x)=3(1)^2+k\sqrt{2}=3+k\sqrt{2}.$

For $x\ge 1$, $f(1)=m(1)^2+k^2=m+k^2.$

Continuity gives 3+k\sqrt{2}=m+k^2. \tag{1}

3. Differentiability at $x=1$

Left derivative: $\frac{d}{dx}\left(3x^2+k\sqrt{x+1}\right)=6x+\frac{k}{2\sqrt{x+1}}$ Hence, $f'_-(1)=6+\frac{k}{2\sqrt{2}}.$

Right derivative: $\frac{d}{dx}(mx^2+k^2)=2mx$ Hence, $f'_+(1)=2m.$

Differentiability gives 6+\frac{k}{2\sqrt{2}}=2m. \tag{2}

4. Solve for $k$ and $m$.

From (2), $m=3+\frac{k}{4\sqrt{2}}.$

Substitute into (1): $3+k\sqrt{2}=\left(3+\frac{k}{4\sqrt{2}}\right)+k^2.$ Cancel $3$: $k\sqrt{2}=\frac{k}{4\sqrt{2}}+k^2.$ Since $k>0$, divide by $k$: $\sqrt{2}=\frac{1}{4\sqrt{2}}+k.$ Now, $\frac{1}{4\sqrt{2}}=\frac{\sqrt{2}}{8},$ so $k=\sqrt{2}-\frac{\sqrt{2}}{8}=\frac{7\sqrt{2}}{8}.$

Then $m=3+\frac{k}{4\sqrt{2}}=3+\frac{7\sqrt{2}/8}{4\sqrt{2}}=3+\frac{7}{32}=\frac{103}{32}.$

5. Compute $f'(8)$ and $f'\left(\frac18\right)$.

Since $8\ge 1$, use the second branch: $f'(x)=2mx$ so $f'(8)=2m\cdot 8=16m=16\cdot \frac{103}{32}=\frac{103}{2}.$

Since $\frac18<1$, use the first branch: $f'(x)=6x+\frac{k}{2\sqrt{x+1}}.$ Thus $f'\left(\frac18\right)=6\cdot \frac18+\frac{k}{2\sqrt{1+1/8}}$ $=\frac34+\frac{k}{2\sqrt{9/8}}.$ Now, $\sqrt{\frac98}=\frac{3}{2\sqrt{2}},$ so $2\sqrt{\frac98}=\frac{3}{\sqrt{2}}.$ Hence $\frac{k}{2\sqrt{9/8}}=k\cdot \frac{\sqrt{2}}{3}.$ With $k=\frac{7\sqrt{2}}{8}$, $k\cdot \frac{\sqrt{2}}{3}=\frac{7\cdot 2}{24}=\frac{7}{12}.$ Therefore, $f'\left(\frac18\right)=\frac34+\frac{7}{12}=\frac{9+7}{12}=\frac{16}{12}=\frac43.$

6. Now evaluate $\frac{8f'(8)}{f'(1/8)}=\frac{8\cdot (103/2)}{4/3}=412\cdot \frac34=309.$

Therefore, the required integer is $\boxed{309}.$