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)}`

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@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}`.