A noiseless channel can carry an arbitrarily large amount of information, no matterhow often it is sampled. Just send a lot of data per sample.
now for second case Shannon limit apply here
Shannon's Theorem
Shannon's Theorem gives an upper bound to the capacity of a link, in bits per second (bps), as a function of the available bandwidth and the signal-to-noise ratio of the link.
The Theorem can be stated as:
C = B * log2(1+ S/N)
where C is the achievable channel capacity, B is the bandwidth of the line, S is the average signal power and N is the average noise power.
The signal-to-noise ratio (S/N) is usually expressed in decibels (dB) given by the formula:
10 * log10(S/N)
but S/N is given 30 dB (in decibal)
so
10 * log10(S/N) = 30 dB.
so S/N=1000
given 4khz
so
C = 4000 * log2(1001)=39.86kbps (shannon'theorem )
Theory : http://www.inf.fu-berlin.de/lehre/WS01/19548-U/shannon.html