Answer: \(\boxed{\dfrac{671}{1296}}\) and \(\boxed{N=18}\).

Each throw has probability

\[p=\frac16\]

of obtaining a 6. The probability of not obtaining a 6 on a throw is

\[q=\frac56.\]

(i) Obtaining a 6 in no more than four throws is the complement of getting no 6 in the first four throws:

\[P(X\le 4)=1-q^4.\]

So

\[P(X\le 4)=1-\left(\frac56\right)^4.\]

Therefore

\[P(X\le 4)=1-\frac{625}{1296}=\boxed{\frac{671}{1296}}.\]

(ii) “Before the \(N\)th throw” means during the first \(N-1\) throws. Hence

\[1-\left(\frac56\right)^{N-1}>0.95.\]

So

\[\left(\frac56\right)^{N-1}<0.05.\]

Taking logarithms,

\[(N-1)\log\left(\frac56\right)<\log(0.05).\]

Since \(\log(5/6)\) is negative, the inequality reverses when we divide:

\[N-1>\frac{\log(0.05)}{\log(5/6)}.\]

Now

\[\frac{\log(0.05)}{\log(5/6)}=16.43\ldots\]

Therefore

\[N-1\ge 17,\]

so the least possible integer is

\[\boxed{N=18}.\]