The wavelength of a travelling wave is 4m at a frequency of 10hertz. what time later will there be another crest at x=3m? ii) if the amplitude of the wave is 12m, write the equation of the wave?

I would write the equation as:
#y(x,t)=Acos(kx-omegat)#
where:
#A=#amplitude;
#k=(2pi)/lambda# is the wavenumber (#lambda# is the wavelength);
#omega=(2pi)/T# is the angular frequency (#T# is the period).
You have everything here:
From the frquency you get the period as:
#T=1/f=1/10=0.1s# that incidentally is the time to have the next crest at your position (the period represents the time to complete an oscillation).
#lambda=4m#
so finally your wave (propagating in the #+x# direction) will be:
#color(red)(y(x,t)=12cos(1.57x-62.83t))#

A basic wave@ may be expressed in the following general equation
#psi (x,t) = A e^{i( kx - \omega t)} # ..........(1)
where #A# is the maximum amplitude,
wavenumber #k=(2pi)/lambda#, and
angular frequency #omega=2pif# .
(#lambda# is wavelength and #f# is the frequency of the wave).

From the given frequency of #10Hz#, we get time period #T=1/f#
#T=1/10=0.1s#. Also given is #lambda=4m#
(i) What time later will there be another crest at #x=3m#.
We know that for any value of #x#, the traveling wave repeats itself as per time-period of the wave. Therefore, if crest appeared at #t=0#; the next crest will occur at
#t=0+T=0+0.1=0.1s#

(ii) Inserting values of #A,k and omega # in (1) we obtain
#psi (x,t) = 12 e^{i( (2pi)/4x - 2pi xx10 t)} #
#psi (x,t) = 12 e^{i( (pi)/2x - 20pi t)} #

.-.-.-.-.-.-.-.-.-.-.-.-.--.-.-.-.-.-.-.-.-.-

@which can be shown to be the usual sine and cosine forms using Euler's formula.
Rewriting the argument as,
#kx-\omega t =(2pi)/lambdak-2pift#, after substituting #v=flambda#
#= (\frac{2\pi}{\lambda})(x - vt)#
We see that this expression describes a vibration of wavelength #\lambda = \frac{2\pi}{k}# traveling along the #x#-direction with a constant phase velocity given by #v_p = \frac{\omega}{k}#.