Convolution sum vs auto-correlation problem
There are two things to consider. First, looking only at the exam question you can say that the convolution of $h[n]$ with a sequence $y[n]$ is given by

$$(h*y)[k]=\sum_nh[n]y[k-n]$$

Now define $y[n]=x[-n]$:

$$(h*y)[k]=\sum_nh[n]x[n-k]$$

Consequently, we have

$$\sum_nh[n]x[n+k]=(h*y)[-k],\quad y[n]=x[-n]$$

So you need to flip the signal $x[n]$, compute the convolution, and flip the result. This is true, given the sum in your exam question. No need to worry about complex conjugates because we just used the definition of the autocorrelation sum in the question.

The other issue is the definition of the autocorrelation for complex sequences, and how it can be implemented by convolution. The usual definition is

$$R_{xx}[k]=\sum_nx^{*}[n]x[n+k]$$

The convolution of $x^*[n]$ with a sequence $h[n]$ is

$$(x^**h)[k]=\sum_nx^*[n]h[k-n]$$

For $h[n]=x[-n]$ we have

$$(x^**h)[k]=\sum_nx^*[n]x[n-k]=R_{xx}[-k]$$

So for complex signals, you need to compute the convolution sum of $x^*[n]$ with $x[-n]$, and flip the result to arrive at the autocorrelation.

Note that $R_{xx}[k]=R_{xx}^*[-k]$, so instead of flipping the result, you can also use its complex conjugate.